tag:blogger.com,1999:blog-621741708273407532024-03-13T06:16:03.814-07:00IIT JEE and NEET PhysicsA Complete Physics Resource for preparing IIT-JEE,NEET,CBSE,ICSE and IGCSE. All Physics topics are divided into multiple sub topics and are explained in detail using concept videos and synopsis.Lots of problems in each topic are solved to understand the concepts clearly.Unknownnoreply@blogger.comBlogger372125tag:blogger.com,1999:blog-62174170827340753.post-65243115759516827742017-04-22T22:43:00.000-07:002017-04-22T22:43:15.417-07:00Electric Intensity and Torque due to Electric Dipole Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Electric dipole is the combination of two charges of equal magnitude but opposite nature separated by small distance.Electric intensity is the force experienced by a unit positive charge placed in the electric field of dipole. Torque is the turning experience got by a electric dipole when placed in an external electric field.</span></div>
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<span style="color: blue; font-size: large;"><b>Electric Field intensity on the axial line of dipole</b></span><br />
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<span style="font-size: large;">A line passing through two charges is called axial line. We would like to measure the electric field intensity at any point on the axial line of electric dipole. Let us consider a point on the axial line at a distance r from the center of the dipole. We shall imagine a unit positive charge at the considered point. It experience force both due to positive and negative charge. Due to Positive charge force is repulsive and due to negative charge it is attractive. Using Coulomb's inverse square law, we need to write equations for the force experienced by unit positive charge at the given location.We need to measure the effective force as the difference between the two charges and it can be further simplified as shown in the video lesson below. Electric dipole moment is the product of any one charge of the dipole to the distance between the two charges of dipole. It is a vector quantity and its direction is from negative charge towards positive charge. Intensity is expressed in terms of dipole moment as shown below.</span></div>
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<span style="color: blue; font-size: large;"><b>Electric field intensity on equatorial line of Dipole</b></span></center>
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<span style="font-size: large;">Equatorial line is a line passing through the center of dipole and perpendicular to the axial line. Let us consider a point on that line that is at a finite distance from center of dipole. We shall imagine a unit positive charge at that point and it experience force due to both positive and negative charge of the dipole. Its magnitude can be determined using inverse square law and its value is shown in the video below. The force due to positive charge is repulsive on unit positive charge and force due to negative charge is attractive. Their directions were identified and the resultant is determined using the vector laws of addition as shown in the video lesson below. It can be noticed that the electric intensity on the equatorial line is half that of intensity on the axial line. A detailed proof is given in the video lesson below. Its direction is also shown.</span></center>
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<span style="color: blue; font-size: large;"><b>Electric Intensity at any point on the dipole</b></span></center>
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<span style="font-size: large;">Let us consider a point around the dipole that is neither on the axial line or equatorial line and the point is at a distance and is making some angle with the horizontal line. To find the electric field intensity at that point, we can consider the dipole as the combination of two dipoles that are perpendicular to each other as shown in the video lesson. For one dipole the considered point is on the axial line and for the other imagined dipole the point is on the equatorial line. As we have derived the equations for the intensity on axial line and equatorial line, we can use that equations and they two are perpendicular to each other. By simplifying them further as shown in the video lesson, we can get the resultant equation as shown. This is a generic equation and in that equation, if the angle is zero, the point will be on axial line and if the angle is ninety degree, the point goes to equatorial line. </span></center>
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<span style="font-size: large;">Let us consider a dipole of two charges separated by a small distance and let us apply a electric field of known intensity on it. Each charge experience a force and and the two forces are equal in magnitude but opposite in direction. But they don't cancel each other as the two forces are acting on different points of the electric dipole. Thus it experience a turning effect in anti clock wise direction and we can measure the torque as shown in the video lesson below. Torque is defined as the product of any one force and the perpendicular distance between the two forces acting on the dipole. It is a vector and we can find the direction using the right hand thumb rule or cork screw rule as shown in the video lesson below.</span></center>
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<span style="font-size: large;">P<a href="http://www.venkatsacademy.com/2017/03/properties-of-electric-charges-video-lesson.html" style="color: #888888; text-decoration-line: none;">roperties of Electric Charges Video Lesson</a></span></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-76186180129071168472017-03-26T07:39:00.004-07:002017-04-22T22:45:03.353-07:00Neutral Point and Equilibrium of Electric charge in a line Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Electrically neutral point is a location at which the resultant electric force is zero. It means the charge kept at that point is experiencing equal and opposite force due to the other two charges kept in that line. To find the neutral point, we can equate the two forces and when the two charges are of similar nature, we can get null point between the charges existing in one line and beyond them, we are not going to find the neutral point. At a location beyond the neutral point, the force acting on the third charge are in the same direction and hence there will not be any neutral point. It is explained in the following video lesson and the location of neutral point is also derived.</span></div>
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<span style="font-size: large;">When the charges are of opposite nature, the resultant force on a third charge in between them is not zero and it is experiencing the two force in the same directions. If we consider a location out side the two charges and consider a third charge, it experience forces due to two charges in opposite direction. If they are equal in magnitude, we can get null point as shown in the video lesson. The expression for the location of null point is also derived here.</span></center>
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<span style="color: blue; font-size: large;"><b>Equilibrium of system of three charges</b></span></center>
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<span style="font-size: large;">Now we are considering a scenario of three charges kept on a straight line. If the two charges at the ends of the line, we can get null point between them weather we keep positive or negative charges in between them. But to keep any of the positive charge at the end of the system, we shall only keep a negative charge but not positive charge in between them. Thus to keep system of three charges in equilibrium, we need to keep two positive charges at the ends and a negative charge in between. We need to write the conditions for zero resultant force location in at least on two charges so that we can get the magnitude,location and nature of the charge in between to get the answer as shown in the video lesson below.</span></center>
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<b style="color: blue; font-size: x-large;">Finding separation between two charges suspended from rigid support</b></center>
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<span style="font-size: large;">Let us consider two identical ball of same mass and similar charge kept suspended from a rigid support using a light wire and because of the repulsion there will be some angular as well as physical separation between them. We need to find that angular separation in terms of the given data and the video solution is as shown below.</span></center>
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<span style="font-size: large;">Let us consider a charged particle is moving horizontally with some constant velocity and an electric field is applied perpendicular to the motion. Gravitational force is small and we can ignore it in this case. Thus there is no force in the horizontal direction and hence the charged particle will have constant velocity over the horizontal direction. </span></center>
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<span style="font-size: large;">There is no initial velocity along the vertical direction but there is electric force due to intensity along the Y axis. Thus the particle experience acceleration along Y axis and hence the velocity of the particle increases on that axis. We can find the final velocity along each direction and its displacement along both the axes. It is further shown in the video lesson that the path of the charged particle is parabola as shown below.</span></center>
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<span style="color: #888888; text-decoration: none;"><span style="font-size: large;">P<a href="http://www.venkatsacademy.com/2017/03/properties-of-electric-charges-video-lesson.html" style="color: #888888; text-decoration: none;">roperties of Electric Charges Video Lesson</a></span></span></h3>
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<a href="http://www.venkatsacademy.com/2017/04/electric-intensity-and-torque-due-to-dipole-video-lesson.html" style="color: #888888; text-decoration: none;">Electric Intensity and Torque due to Electric Dipole Video Lesson</a></h3>
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Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-62174170827340753.post-3181587246588142552017-03-23T07:54:00.002-07:002017-03-26T07:40:40.542-07:00 Electric force and Coulomb's Inverse square law Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Electric force of attraction or repulsion exists between every two charges and it follows Coulomb's inverse square law. According to the law, the force between two charges is directly proportional to the product of charges and is inversely proportional to the square of the distance of separation.</span></div>
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<span style="font-size: large;">Gravitational and electric forces are fundamental in nature. Gravitational force is due to mass of the particle and it exits between every two masses where as electric force exits between the the particles or bodies having only excess charges. Both of them obeys inverse square law but gravitational force is independent of medium but electric force depends on the medium between charges.</span></div>
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<span style="font-size: large;">Gravitational force is the weakest of fundamental forces but electric force is much stronger than that. Gravitational force is further always attractive force but electric force is either attractive or repulsive. Further classification is explained in the video lesson below.</span></div>
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<span style="font-size: large;">To identify the relation between the force and charges different experiments were conducted and we got to a conclusion that the electric force between the charges is directly proportional to the product of charges and is inversely proportional to the square of the distance of the separation. The force also depends on the medium between the charges and its impact is studied basing on the permitivity of the medium. The ratio of permitivity of the medium to the permitivity of the free space is called dielectric constant or relative permitivity. Its value for the vacuum is one and for any non conducting medium it is greater than one. It is explained as shown in the video lesson below.</span></center>
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<span style="font-size: large;">As we have explained earlier, we can understand that the force between the charges gets reduced when the medium is placed instead of the vacuum. It is further explained in the video lesson below. We can find out the effective distance in the presence of medium as shown below.</span></center>
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<span style="font-size: large;">Electric force between the charges is not effected even if we keep a new charge in between them. What makes a difference is, more forces are acting on each charge and hence resultant force on each charge is more. But the force between two charges is not affected because of the third charge. To find the resultant charge, we can use vector laws of addition. It is explained in the video lesson below.</span></center>
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<span style="color: #888888;"><span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/07/electric-potential-problems-with-solutions.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Electric Potential Problems with Solutions</span></a></span></span></h3>
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<span style="color: #888888; text-decoration: none;"><span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/07/energy-stored-in-capacitor-and-effect-on-energy.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Energy Stored in Capacitor and Effect of Dielectric on it</span></a></span></span></h3>
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/03/properties-of-electric-charges-video-lesson.html" style="color: #888888; text-decoration: none;">Properties of Electric Charges Video Lesson</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/neutral-point-and-equilibrium-of-electric-charges-video-lesson.html" style="color: #888888; text-decoration: none;">Neutral Point and Equilibrium of Electric charge in a line Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-62155927012897878412017-03-20T07:34:00.000-07:002017-03-23T07:55:07.706-07:00Properties of Electric Charges Video Lesson <div style="text-align: justify;">
<span style="font-size: large;">Electric charge is the property of a particle of certain mass due to which it can either attract the opposite charge or repel the similar charge.Charges are classified as positive and negative and many bodies around are neutral which means they have equal and opposite charges. </span></div>
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<span style="font-size: large;">When one body with excess charges is rubbed with the other, there is transfer of electrons from one body to other. The body lost electrons will have excess positive charges and the body that gained electrons will have excess electrons. Thus the body that gained electrons is negatively charged and the body that lost electrons is positively charged.</span></div>
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<span style="font-size: large;">Thus body that is positively charged losses a small portion of mass due to loss of electrons and the body that gained electrons is not only negatively charged but its mass is also slightly increases.</span></div>
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<span style="font-size: large;">Similar charges repel each other and opposite charges attract each other. This force of attraction obeys inverse square law. According to it the force of attraction or repulsion is inversely proportional to the square of distance of separation. It is a long distance range force. Further properties are explained in the video lesson below.</span></center>
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<span style="color: blue; font-size: large;"><b>Number of electrons in one coulomb charge</b></span></center>
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<span style="font-size: large;">We know that electron is treated like a fundamental charge and it is measured in a unit called coulomb. We also know that the charge is always available as integral multiples of charge of electron and this concept is called quantization of the charge. Taking that concept into consideration, we can find the number of electrons in one coulomb charge as shown in the video lesson below.</span></center>
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<b style="color: blue; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><span style="font-size: large;">Methods of electric charging</span></b></center>
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<span style="font-size: large;">We can charge the body by rubbing one body with the other. This is called charging a body by friction and during the rubbing, one body loose the electrons and the other body gains the electrons. We can also charge a body by conducting where we are passing the charge through the materials which has the ability to pass the current. We can also charge a body by keeping the body in contact. In that case excess charges from one body to other get transferred until there is some potential difference between them. It is explained in the following video lesson.</span></center>
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<b style="color: blue; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><span style="font-size: large;">Charging a body by induction</span></b></center>
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<span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif;"><span style="font-size: large;">Induction is the phenomenon of rearrangement of charges in a neutral body in the presence of a charged body. This separation of charges is called polarisation of charges and we can get a body charged using induction method with out getting in touch and contact with the body. It is explained in the following video lesson below.</span></span></center>
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<b style="color: blue; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><span style="font-size: large;">Related Posts</span></b></center>
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<a href="http://www.venkatsacademy.com/2017/03/free-body-diagram-and-connected-bodies-video-lesson.html" style="color: #888888; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; text-align: start; text-decoration: none;"><b><span style="font-size: large;">Free body diagram and Connected Bodies Video Lesson</span></b></a></center>
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<span style="font-size: large;">Electric Charge and Electric Force</span></h3>
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<a href="http://www.venkatsacademy.com/2016/07/resultant-force-and-coloumbs-law-of-electric-charges.html" style="color: #888888; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; line-height: normal; text-align: start; text-decoration: none;"><b><span style="font-size: large;">Resultant Force and Coloumb's Law of Electric Force Problems and Solutions</span></b></a></center>
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<a href="http://www.venkatsacademy.com/2016/07/electric-potential-expression-and-explanation.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Electric Potential Expression and Explanation</span></a></h3>
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<span style="color: #888888;"><span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/07/electric-potential-problems-with-solutions.html" style="color: #888888; text-decoration: none;">Electric Potential Problems with Solutions</a></span></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/electric-force-and-coulombs-inverse-square-law-video-lesson.html" style="color: #888888; text-decoration: none;">Electric force and Coulomb's Inverse square law Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-47861263783054370892017-03-04T21:34:00.003-08:002017-03-04T21:34:45.854-08:00Free body diagram and Connected Bodies Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Free body diagram is the representation of all the forces acting on a body at a given instant. We shall consider only the forces acting on a body but not the forces applied by the same body. When there are multiple bodies, all of them can be treated as one system if they have same acceleration. When we are solving a system of bodies and we are interested in finding the acceleration and may be tension and contact forces on a given body. We shall draw free body diagram for each body and we shall further write the equation for the resultant force. We shall consider the force along the direction of motion as positive and vice versa. We are also dealing with contact force in the given video below. Contact force is t he force applied by one body on the other when they are in contact. There will be reaction force also according to Newton's third law of motion.A detailed video lesson is presented below regarding free body diagrams as shown below.</span></div>
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<span style="font-size: large;">Here we are discussing about the bodies of different masses connected with the help of string and a force is applied on the system. When force is applied, the string becomes tight and a tension is developed in the wire. We need to identify all the forces acting on each body. We need to be careful and identify only forces acting on the body but not the forces applied by the body. This can be done with the help of free body diagram for each body. Thus we can write equation for resultant force on each body and we can solve that equations and get the tension in each wire and acceleration of the system as shown in the video lesson below.</span></center>
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<span style="font-size: large;">Two bodies of different masses connected over a smooth pulley with the help of light weighted string could be called as Atwood's machine. We need to find the tension int the wire and the acceleration of the system. We can write free body diagram for each body and equations for resultant force as shown in the video lesson below.</span></center>
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<span style="font-size: large;">Let us consider a system of two bodies where one body is hanging vertically from a table and the other body is on a smooth horizontal smooth surface and the two bodies are connected with a light weighted string over a smooth pulley. We need to find the tension in the string and the acceleration of the system. We can draw free body diagram to identify the forces acting on each body and we need to write the equations for resultant force as shown in the video lesson below. We can solve the equations as shown below.</span></center>
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<b><span style="color: #888888; line-height: normal; text-align: start; text-decoration: none;"><a href="http://www.venkatsacademy.com/2015/10/newton-law-of-inertia-and-newton-law-of-force.html" style="color: #888888; line-height: normal; text-align: start; text-decoration: none;"><span style="font-size: large;">Newton Law of Inertia and Newton law of Force an Introduction</span></a></span></b></center>
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<b><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-one.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions One</a></b></div>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-two.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Two</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-three.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Three</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-FOUR.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions four</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-five.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Five</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/linear-momentum-and-conservation-of-momentum-video-lesson.html" style="color: #888888; text-decoration: none;">Linear Momentum and Conservation of Momentum Video Lesson</a></h3>
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<a href="http://www.venkatsacademy.com/2017/03/newtons-first-and-second-law-of-motion-video-lesson.html" style="color: #888888; text-decoration: none;">Newton's First and Second Law of Motion Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-21485883083724111262017-03-04T20:35:00.001-08:002017-03-04T21:36:18.538-08:00Newton's First and Second Law of Motion Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Newton's first and second law are helpful in understanding translatory motion. Newton's first law explains the concept of inertia. Any body has the property of continuing its own state of rest or motion when no external force is acting on the body. This property is called inertia. This itself is called first law of motion. Body can have inertia of rest,motion and direction.</span></div>
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<span style="font-size: large;">Inertia of rest is the property of body due to which the body continue its state of rest until no external force is acting on the body. We can experience it in daily life. Consider a person in standing position that is in the </span><span style="font-size: large;">in a vehicle that is at rest.</span><span style="font-size: large;">When the bus suddenly starts, the connected part of the person with the vehicle that are legs starts moving. But the upper part of the body tries to continue in the state of rest due to inertia of rest. Thus the lower part of the body moves and the upper part stays there it self. So he fell like he got a jerk and fell like falling back. it is due to inertia of rest. The same can be explained with inertia of rest and direction. A Video lesson about the first law is shown below for your reference.</span></div>
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<span style="font-size: large;"><br /></span><span style="color: blue; font-size: large;"><b>Newton's Second Law of motion</b></span><br />
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<span style="font-size: large;">Newton's second law of motion is about force. Force is a physical quantity that changes or tries to change the state of the body. Force is defined as the rate of change of momentum. If a body is having a constant mass, then force can be defined as the product of mass and acceleration of the body. Force is a vector that has both magnitude and direction and it is a vector quantity. The direction of the force is similar to the acceleration. It is explained in the video lesson below.</span><br />
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<span style="font-size: large;">When there is a resultant force acting on a body, then the body velocity gets changed and hence it will have some acceleration. If there are multiple force acting on a body, we shall find the effective force or resultant force using vector laws of addition. If there is no resultant force acting on a body, the body will be either moving with a constant velocity or in the state of rest basing on its previous condition. We can find the resultant force acting on a body as shown in the video lesson below.</span></center>
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<b style="background-color: white; color: blue; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><span style="font-size: large;">Related Posts</span></b></center>
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<b style="color: #666666; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><a href="http://www.venkatsacademy.com/2015/10/newton-law-of-inertia-and-newton-law-of-force.html" style="color: #888888; line-height: normal; text-align: start; text-decoration: none;"><span style="font-size: large;">Newton Law of Inertia and Newton law of Force an Introduction</span></a></b></center>
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<b><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-one.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions One</a></b></div>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-two.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Two</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-three.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Three</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-FOUR.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions four</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-five.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Five</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-six.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Six</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-sevven.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Seven</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-eight.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Eight</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/linear-momentum-and-conservation-of-momentum-video-lesson.html" style="color: #888888; text-decoration: none;">Linear Momentum and Conservation of Momentum Video Lesson</a></b></h3>
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<a href="http://www.venkatsacademy.com/2017/03/free-body-diagram-and-connected-bodies-video-lesson.html" style="color: #888888; text-decoration: none;">Free body diagram and Connected Bodies Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-71981810530761473982017-03-04T19:52:00.002-08:002017-03-04T20:36:13.866-08:00Linear Momentum and Conservation of Momentum Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Linear momentum is a physical quantity that explains about the transfer of kinetic energy of one body with another when there is a collision with other body. It is mathematically defined as the product of mass and velocity of the body. If momentum is more, it transfer kinetic energy to other body more and vice versa. Momentum is a vector quantity and its direction is similar to the direction of velocity.</span></div>
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<span style="font-size: large;">Linear momentum is a physical quantity that explains the translatory motion. We can define Newton's second law in terms of linear momentum. As per the law, force is defined as the rate of change of linear momentum. Momentum is a physical quantity that gives clarity regarding the ability of transfer of kinetic energy. We can also find the relation between linear momentum and kinetic energy and it is explained in the video lesson below.</span></div>
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<span style="font-size: large;">When a rigid body of constant mass changes its velocity, its linear velocity also changes. When a body either change its magnitude or direction of the velocity of the vector or both of them, then there is a change in the linear momentum. We can find the change in the linear momentum of a body as the change in the final linear momentum of the body to its initial linear momentum in a given time. If the body is moving with a constant velocity, then there is no change in the momentum. We can find the change in the momentum in different cases and we can also find the direction.</span></center>
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<span style="font-size: large;">We shall consider the component of the momentum that is actually changing and if any component of the momentum is not changing, then we need not consider it. Some simple cases of change in momentum is explained in the video lesson below.</span></center>
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<span style="font-size: large;">Linear momentum for a body is conserved when no external force is acting on the body. Proving this is very simple. We have defined force as the rate of change of momentum. When applied force is zero, rate of change in momentum is zero and it means momentum is not changing with time. Thus linear momentum remains constant when no external force is acting on a body. It is explained in the video lesson below.</span></center>
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<b><span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-one.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions One</a></span></b></div>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-two.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Two</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-three.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Three</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-FOUR.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions four</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-five.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Five</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-six.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Six</a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-sevven.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Seven</a></span></h3>
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<span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/11/laws-of-motion-problems-with-solutions-eight.html" style="color: #888888; text-decoration: none;">Laws of Motion Problems with Solutions Eight</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/newtons-first-and-second-law-of-motion-video-lesson.html" style="color: #888888; text-decoration: none;">Newton's First and Second Law of Motion Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-27994208195654164252017-03-04T09:06:00.002-08:002017-03-04T09:06:41.739-08:00Horizontal Projectile and its Velocity Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">A body projected from a certain height horizontally with some initial velocity is called horizontal projectile. It has only initial velocity along the horizontal direction but it has no initial velocity along the vertical direction. </span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">But as the body start moving, acceleration due to gravity starts acting, velocity of the body in the horizontal direction remains same as there is no acceleration due to gravity remains same. But velocity component of the body along the Y direction starts increasing. We can find the displacement along the horizontal and vertical directions using the basic equations of motion and we can show that the path is parabola as shown in the video lesson below.</span></div>
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<span style="font-size: large;"><br /></span>
<span style="color: blue; font-size: large;"><b>Path of horizontal Projectile is parabola</b></span></div>
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<span style="color: blue; font-size: large;"><b>Velocity of horizontal projectile</b></span></center>
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<span style="font-size: large;">The horizontal projectile has initial velocity of projection in the horizontal direction but it has no initial velocity along the Y direction. The velocity of projectile in the horizontal direction remains same as there is no acceleration in that direction and the component of the velocity along the Y direction starts increasing. We can find the final velocity of projectile at any instant is the vector sum of horizontal and vertical components of velocity. The final velocity has a certain direction and we can find the direction as shown in the video lecture below.</span></center>
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<span style="color: blue; font-size: large;"><b>Angle of projectile of oblique projectile where range and maximum heights are same </b></span></center>
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<span style="font-size: large;">A oblique projectile is a body that is projected from the horizontal direction with a certain angle. The range of the projectile is the maximum horizontal distance travelled by a projectile body and the maximum height is the maximum vertical distance that the projectile has travelled. If these two has to he the same, we need to equate the mathematical equations of both of them and we can find the corresponding angle of projection can be measured as shown in the diagram below. </span></center>
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<b style="color: blue;"><span style="font-size: large;">Related Posts</span></b></center>
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<b style="line-height: normal; text-align: start;"><span style="color: blue;"><span style="font-size: large;"><a href="http://www.venkatsacademy.com/2015/10/projectile-motion-path-is-parabola-and-range.html" style="color: #888888; text-decoration: none;">Projectile Motion Range,Time of Flight and Maximum Height Equations</a> </span></span></b></center>
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<span style="color: blue; line-height: normal;"><b><a href="http://www.venkatsacademy.com/2015/10/velocity-of-projectile-and-problems-on-projectile.html" style="color: #888888; line-height: normal; text-decoration: none;"><span style="font-size: large;">Velocity of Projectile and Problems on Projectile Motion</span></a></b></span></div>
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<span style="color: blue;"><span style="color: #888888;"><a href="http://www.venkatsacademy.com/2015/10/horizontal-projectile-applications-and-problems-SOLUTIONS.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Horizontal Projectile, Applications Problems with Solutions</span></a></span></span></h3>
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<span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/03/motion-of-body-in-plane-and-projectile-motion-video-lesson.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion of body in a Plane and Projectile motion Video Lesson</span></a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/range-and-maximum-height-of-projectile-video-lesson.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Range and Maximum height of Projectile Motion Video Lesson</span></a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-90252699490644549642017-03-04T08:43:00.002-08:002017-03-04T09:07:34.882-08:00Range and Maximum height of Projectile Motion Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Range of the projectile is the maximum horizontal distance travelled by a projectile during its time of flight. By the time the end of time of flight, the vertical displacement of the projectile is zero but its initial and final positions along the horizontal direction are different. We can find the horizontal displacement of a projectile as the product of velocity of projectile along the horizontal direction and time taken for the journey. The time taken to complete projectile motion is called time of flight. The horizontal displacement that the projectile has during the time of flight is the maximum horizontal displacement and it is called range of the projectile. We can find the range of the projectile as shown in the video lesson below.</span></div>
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<span style="font-size: large;"><br /></span></center>
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<span style="font-size: large;">It is found that the range of a projectile depends on the initial velocity of projection and angle of projection. So it is clear that the range of a projectile is maximum for an angle of projection 45 degree. It is the reason due to which people try to project any body with an angle close to 45 degree when they want to get maximum horizontal distance like disc and javelin throw.</span></center>
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<span style="color: blue; font-size: large;"><b>Range is same for two angles of projection</b></span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="font-size: large;">It can be proved that the range of the projectile is same for complimentary angels of projection. The heights that they reach is different but the maximum horizontal range that both of them takes for complimentary angles are same. It is shown in video lectures as shown in the video below.</span></center>
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<span style="color: blue; font-size: large;"><b>Maximum height of projectile</b></span></center>
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<span style="font-size: large;">Maximum height of the projectile is the maximum vertical displacement that a projectile has during its journey. We can measure it using the equation of motion of one dimension itself. The projectile has initial velocity component along the vertical direction. As it is moving up against the gravity, its velocity keeps on decreasing and the point at which the vertical component of velocity becomes zero is called maximum direction. We can derive the equation in the video lesson as shown in the diagram below.</span></center>
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<b style="color: blue;"><span style="font-size: large;">Related Posts</span></b></center>
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<b style="line-height: normal; text-align: start;"><span style="color: blue;"><span style="font-size: large;"><a href="http://www.venkatsacademy.com/2015/10/projectile-motion-path-is-parabola-and-range.html" style="color: #888888; text-decoration: none;">Projectile Motion Range,Time of Flight and Maximum Height Equations</a> </span></span></b></center>
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<span style="color: blue; line-height: normal;"><b><a href="http://www.venkatsacademy.com/2015/10/velocity-of-projectile-and-problems-on-projectile.html" style="color: #888888; line-height: normal; text-decoration: none;"><span style="font-size: large;">Velocity of Projectile and Problems on Projectile Motion</span></a></b></span></div>
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<span style="color: blue;"><span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2015/10/horizontal-projectile-applications-and-problems-SOLUTIONS.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Horizontal Projectile, Applications Problems with Solutions</span></a></span></span></h3>
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/03/motion-of-body-in-plane-and-projectile-motion-video-lesson.html" style="color: #888888; text-decoration: none;">Motion of body in a Plane and Projectile motion Video Lesson</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/horizontal-projectile-and-its-velocity-video-lesson.html" style="color: #888888; text-decoration: none;">Horizontal Projectile and its Velocity Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-69242473997464906882017-03-04T07:51:00.002-08:002017-03-04T08:44:13.359-08:00Motion of body in a Plane and Projectile motion Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Any body moving in a plane where it has motion along horizontal and vertical directions, then the motion is called motion in a plane and it is also called two dimensional motion or projectile motion. For a body to have projectile motion, it shall be projected from the ground with an angle other than ninety degree with a certain initial velocity.The body has sentimentally velocity along both the directions. To find the velocity along a given direction, we need to resolve the initial velocity has to be resolved into components. Some component is along the horizontal axis and as there is no gravity along that direction and hence velocity component along the horizontal direction remains constant. The velocity along the vertical directions changes as acceleration due to gravity acts against initial vertical direction and hence the velocity keep changing along that direction. We can find the displacement along the horizontal and vertical directions. It is explained in the video lesson as shown in the video below.</span></div>
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<span style="font-size: large;"><br /></span></center>
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<span style="color: blue; font-size: large;"><b>Path of projectile is parabola</b></span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="font-size: large;">When a body is projected with an angle to the horizontal, it has displacement along both horizontal and vertical directions. We can find them using the equation of motion as shown in the diagram below. The displacement along X direction is independent of acceleration due to gravity as it is acting only along Y direction. We can find the time taken to have horizontal direction in terms of initial velocity as shown in the video lesson below. We can find the displacement along Y direction and we need to substitute the time from the first case and hence we will get a vertical displacement format as mathematical parabola.</span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="color: blue; font-size: large;"><b>Time of flight of Projectile</b></span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="font-size: large;">The time taken by the projectile to reach the ground from the point of projection is called time of flight. We need to know that by the time of end of journey, it comes to some other point on the ground and hence it has some horizontal displacement. But its initial and final position with respect to the Y direction and hence the vertical displacement by the end of time of flight along the vertical direction is zero. Taking this into consideration, we can find the time of flight as shown in the video lesson below.</span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="color: blue; font-size: large;"><b>Velocity of Projectile at any instant</b></span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="font-size: large;">The projectile has some initial velocity and it can be resolved into components along horizontal and vertical direction. As the time progresses, the total velocity of the projectile changes. The horizontal component of velocity of the projectile remains constant but the vertical component of the velocity keeps on changing. To find the final velocity, we need to add final velocity components of X and Y directions and it can be done as shown in the video lesson below.</span></center>
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<b style="background-color: white; color: blue; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><span style="font-size: large;">Related Posts</span></b></center>
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<a href="http://www.venkatsacademy.com/2017/03/dot-product-and-cross-product-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Dot Product and Cross Product of Vectors Video Lesson</span></a></h3>
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<b style="line-height: normal;"><span style="color: blue;"><span style="font-size: large;"><a href="http://www.venkatsacademy.com/2015/10/projectile-motion-path-is-parabola-and-range.html" style="color: #888888; text-decoration: none;">Projectile Motion Range,Time of Flight and Maximum Height Equations</a> </span></span></b></div>
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<span style="color: blue; line-height: normal;"><b><a href="http://www.venkatsacademy.com/2015/10/velocity-of-projectile-and-problems-on-projectile.html" style="color: #888888; line-height: normal; text-decoration: none;"><span style="font-size: large;">Velocity of Projectile and Problems on Projectile Motion</span></a></b></span></div>
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<span style="color: blue;"><span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2015/10/horizontal-projectile-applications-and-problems-SOLUTIONS.html" style="color: #888888; text-decoration: none;">Horizontal Projectile, Applications Problems with Solutions</a></span></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/range-and-maximum-height-of-projectile-video-lesson.html" style="color: #888888; text-decoration: none;">Range and Maximum height of Projectile Motion Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-19363457961540764202017-03-02T08:58:00.003-08:002017-03-04T08:44:51.575-08:00Dot Product and Cross Product of Vectors Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">Dot product and cross product of two vectors is a way of multiplication of the vectors.</span><span style="font-size: large;">Vector is a physical quantity that has both magnitude and direction and satisfies the vector algebra. Some physical quantities demands both magnitude and direction with out which we cannot explain the physical quantity</span><span style="font-size: large;">. Vector has to be added to the other vectors but we can not add one vector with the other vector. Vector subtraction also follows the same rules of vector addition with the concept of negative vector. </span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">Vector quantity can be multiplied with a scalar also and the resultant is a vector with the same magnitude of the given vector. A vector can be multiplied with a vector and the resultant can be a scalar or vector. If two vectors are multiplied and the product is a scalar and that kind of multiplication is called scalar product or dot product of two vectors. To get the dot product of two three dimensional vectors, we shall multiply the corresponding components and add all of them. A video lesson is given below to explain the dot product as shown in the diagram below.</span></div>
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<iframe allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/r_u71xiyFWk" width="560"></iframe></center>
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<span style="color: blue; font-size: large;"><b>Work done as dot product</b></span></center>
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<span style="font-size: large;"><br /></span></center>
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<span style="font-size: large;">We know that work is said to be done when applied force is able to produce some displacement. When we apply some force at some angle on a body on a horizontal surface. The total force is not acting along the horizontal surface and only a component of force is acting along the horizontal. To know that value, we need to resolve the force vector into components. The component of the force acting along the direction of the displacement is producing the displacement and the other component has no impact on the displacement. Thus we need to consider only a component of force and that shall be multiplied with the displacement to get the work done. It is explained as the dot product of force and displacement as shown in the video lesson below.</span></center>
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<span style="font-size: large;">If we multiply one vector with another vector and the resultant of the product is a vector then the kind of vector product is called cross product of the given vectors. The resultant will have not only magnitude and also has the direction. To find the cross product of two vectors of two vectors, we need to multiply the cross product of unit vectors and get the resultant of two unit vectors also a unit vector but having the different direction. To get the direction of cross product of the two vectors we need to use right hand thumb rule or cork screw rule as explained in the video lesson below.</span></center>
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<span style="font-size: large;">Torque means the turning effect and it is a physical quantity useful in understanding the rotational motion of a body. We shall apply a force away from axis of rotation. We know that the rotating particle is at a distance from the axis of rotation. Applied force and the perpendicular distance from the axis of rotation both are vectors. The resultant of two vectors product is also a vector and we can find the torque as shown in the video lesson below.</span></div>
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<b><span style="font-size: large;"><span style="color: blue;"><a href="http://venkatsacademy.blogspot.in/2014/10/vectors-and-their-usage-in-physics.html" style="color: #888888; text-decoration: none;" target="_blank">Vectors and their usage in Physics</a></span></span></b><br />
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<b><span style="font-size: large;"><span style="color: blue;"><a href="http://venkatsacademy.blogspot.in/2014/10/vectors-parallelogram-lawtriangle-law.html" style="color: #888888; text-decoration: none;">Parallelogram law,Triangle law and applications</a></span></span></b><br />
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<b><span style="color: blue; line-height: 18.48px; text-align: justify;"><a href="http://venkatsacademy.blogspot.in/2014/10/scalar-product-and-vector-product-of.html" style="color: #888888; line-height: 18.48px; text-decoration: none;"><span style="font-size: large;">Scalar product and vector product</span></a></span></b></div>
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<span style="color: black;"><a href="http://www.venkatsacademy.com/2016/10/vectors-problems-and-solutions-one.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions One</span></a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-two.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Two</span></a></span></h3>
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<span style="color: #33aaff;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-three.html" style="color: #33aaff; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Three</span></a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-four.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Four</span></a></span></h3>
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<span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/02/vector-resolution-and-laws-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vector resolution and Laws of Vectors Video lesson</span></a></span></h3>
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/02/parallelogram-law-and-addition-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;">Parallelogram law and addition of vectors Video Lesson</a></span></h3>
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Motion of body in a Plane and Projectile motion Video Lesson</h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-43683049197082230152017-02-22T05:42:00.003-08:002017-03-02T09:00:36.091-08:00Parallelogram law and addition of vectors Video Lesson<div style="text-align: justify;">
<span style="font-size: large;">We are going to deal about parallelogram law of vectors and using the law to find the addition and subtraction of the given vectors. Vector is a physical quantity that has both magnitude and direction. When we add scalars, we only need to worry about their magnitude, But adding vectors is little complicated when compared with scalars. We can add them using a basic graphical method. Here we need to shift the second vector in parallel so that the magnitude and direction remains same. Then the tail of the first vector has to be joined with head of second vector to get the resultant of the two vectors. This is little graphical and performing this method is lengthy process when multiple vectors are involved.</span></div>
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<span style="font-size: large;">The alternate method to add the two vectors is algebric method using parallelogram law of vectors. According to the law, if two vectors are represented as two adjusent sides of a parallelogram starting from the same point, then the resultant of the two vectors is the diagonal of the parallelogram and its direction also can be found as shown in the video below.</span></div>
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<span style="font-size: large;"><b>W</b>e can apply the above mentioned law for different cases. What will be the resultant of the two vectors depends on the magnitude of the two vectors, the angle between them. Here in the below video, we are solving different basic possible cases and we have found the resultant of the given two vectors using the parallelogram law of the vectors.</span></center>
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<span style="font-size: large;">We also would like to consider the addition of the two vectors using algebric method. Here we represent the vectors with components along the X,Y and The Z axis. When we are adding the vectors, we add the respective components and the find the resultant vector and its direction as shown in the video below.</span></center>
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<span style="color: blue; font-size: large;"><b>Relative Velocity</b></span></center>
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<span style="font-size: large;">Relative velocity is different from resultant velocity. Relative velocity is the comparative velocity of one body with respect to the other. A body will have relative velocity only when it has effective displacement when compared with the other body. In the following video, it is explained the way of measuring the relative velocity in brief.</span></center>
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<span style="font-size: large;"><b><span style="color: blue;"><a href="http://venkatsacademy.blogspot.in/2014/10/vectors-and-their-usage-in-physics.html" style="color: #888888; text-decoration: none;" target="_blank">Vectors and their usage in Physics</a></span></b><br /><b><br /></b></span></div>
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<span style="font-size: large;"><b><span style="color: blue;"><a href="http://venkatsacademy.blogspot.in/2014/10/vectors-parallelogram-lawtriangle-law.html" style="color: #888888; text-decoration: none;">Parallelogram law,Triangle law and applications</a></span></b><br /><b><br /></b></span></div>
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<b><span style="color: blue; line-height: 18.48px; text-align: justify;"><a href="http://venkatsacademy.blogspot.in/2014/10/scalar-product-and-vector-product-of.html" style="color: #888888; line-height: 18.48px; text-decoration: none;"><span style="font-size: large;">Scalar product and vector product</span></a></span></b></div>
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<span style="color: black;"><a href="http://www.venkatsacademy.com/2016/10/vectors-problems-and-solutions-one.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions One</span></a></span></h3>
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<span style="color: #33aaff;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-three.html" style="color: #33aaff; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Three</span></a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-four.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Four</span></a></span></h3>
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/02/vector-resolution-and-laws-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;">Vector resolution and Laws of Vectors Video lesson</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/03/dot-product-and-cross-product-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;">Dot Product and Cross Product of Vectors Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-24675176011335366612017-02-21T07:10:00.003-08:002017-02-22T05:43:40.098-08:00Vector resolution and Laws of Vectors Video lesson<div style="text-align: justify;">
<span style="font-size: large;">We are here discussing regarding the introduction of vectors, resolution of vectors and laws of vectors. Vector is a physical quantity that has magnitude, direction and satisfy the laws of vector algebra. Some physical quantities in physics need both magnitude and direction to explain them completely and that kind of vectors are treated as vectors. Simple examples of vectors are displacement,velocity,momentum,force and torque.</span></div>
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<span style="font-size: large;">Vectors are graphically represented as an arrows and the head of the arrow is treated as its direction. The size of the arrow is directly proportional to the magnitude of the vector. The direction of the vector is identified with its unit vector and it has only direction of the vector and the magnitude of the unit vector is one unit only. Thus any vector can be represented as the product of magnitude of the vector and its unit vector.</span></div>
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<span style="font-size: large;">The direction of the vector are represented as unit vectors i,j and k along X,Y and Z axis. All of they are having a magnitude of one unit. If two vectors are having same magnitude and same direction then the two vectors are called unit vectors. If their magnitude is same but the direction is opposite, then one vector is called negative vector of the other vector and vice versa. Here is a video lesson on the basics of the vectors.</span></div>
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<span style="font-size: large;">A vector could be only along one direction and then identifying the direction is easy like it is along X or Y axis. If any vector is in a plane making some angle with any of the axis, then we can not say that it is either along the X or Y axis. To know how much part of the given vector is along the given axis, we can resolve the vector into components. A component is a part of a given vector along a specified direction and we use trigonometry to resolve the vector into components. If we add the two vectors, we will get back the original vector without any loss. It is explained in the video lesson as shown in the diagram below.</span></center>
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<span style="color: blue; font-size: large;"><b>Laws of vectors</b></span></center>
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<span style="font-size: large;">Vectors addition and subtraction can be done either graphically or algebraically. For graphical addition or subtraction, we are going to shift the given and required vector parallel. As shifting the vector in parallel won't change either its magnitude or direction, the vector remains same. Any way this can be done with much ease using mathematical tools. Vector addition satisfies commutative law, and distributive law. But vector subtraction does not satisfy the commutative law. It is explained as shown in the video lesson below.</span></center>
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<b><span style="font-size: large;"><span style="color: blue;"><a href="http://venkatsacademy.blogspot.in/2014/10/vectors-and-their-usage-in-physics.html" style="color: #888888; text-decoration: none;" target="_blank">Vectors and their usage in Physics</a></span></span></b><br />
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<b><span style="font-size: large;"><span style="color: blue;"><a href="http://venkatsacademy.blogspot.in/2014/10/vectors-parallelogram-lawtriangle-law.html" style="color: #888888; text-decoration: none;">Parallelogram law,Triangle law and applications</a></span></span></b><br />
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<b><span style="font-size: large;"><span style="color: blue; line-height: 18.48px; text-align: justify;"><a href="http://venkatsacademy.blogspot.in/2014/10/scalar-product-and-vector-product-of.html" style="color: #888888; line-height: 18.48px; text-decoration: none;">Scalar product and vector product</a></span></span></b></div>
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<span style="color: black;"><a href="http://www.venkatsacademy.com/2016/10/vectors-problems-and-solutions-one.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions One</span></a></span></h3>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-two.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Two</span></a></span></h3>
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<span style="color: #33aaff;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-three.html" style="color: #33aaff; text-decoration: none;"><span style="font-size: large;">Vectors Problems and Solutions Three</span></a></span></h3>
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/11/vectors-problems-and-solutions-four.html" style="color: #888888; text-decoration: none;">Vectors Problems and Solutions Four</a></span></h3>
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<h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-size: 22px; font-stretch: normal; font-weight: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative; text-align: start;">
<a href="http://www.venkatsacademy.com/2017/02/parallelogram-law-and-addition-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;">Parallelogram law and addition of vectors</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-7777105473647982002017-02-04T18:28:00.001-08:002017-02-21T07:13:08.089-08:00Equations of Motion for Vertically Thrown up body<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Equations of motion for a vertically
thrown up body<o:p></o:p></b></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span style="font-size: large;">Thus the velocity of the body keeps on
decreasing and at a particular height, its velocity becomes zero and that
height is said to be the maximum height that the body can go and the time taken
to reach that maximum height is called time of ascent. </span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">Acceleration in this
case is acceleration due to gravity which is constant and it shall be treated
as negative as the velocity of the body keeps on decreasing with respect to
time. Taking this into consideration, we can rewrite the four equations of
motion for a freely falling body. It is proved that the velocity with you
project the body vertically up is the velocity with which it comes back to the
same point of projection but in the opposite direction. It is also proved that
when air resistance is ignored, time of ascent is equal to the time of descent.<o:p></o:p></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<center>
<iframe allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/NBin3GM5fdM" width="560"></iframe></center>
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b><br /></b></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Time of Ascent and Decent with Air resistance</b></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">As it is mentioned earlier time of
ascent of a vertically thrown up body is equal to the time of descent of a
freely falling body. The above mentioned statement is true only when air
resistance is ignored. But in real life air resistance is there and if that is
taken into consideration, time of ascent is different from time of descent. Air
resistance offers a force against the motion and it always acts opposite the
motion. Thus when the body is thrown up, both gravitational force and resistance
force acts against the motion. When the body is coming down, gravitational
force acts in down ward direction but air resistance acts opposite to the
motion and that is in upward direction. Thus we can find the effective
acceleration in both the cases and find the the relations between time of
ascent and descent as shown in the video below.<o:p></o:p></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<br />
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<center>
<iframe allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/ANPSBgmGIcQ" width="560"></iframe></center>
<br />
<div style="background-color: white; color: #666666; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; line-height: 18.48px;">
<span style="color: blue;"><b><span style="font-size: medium;"><br /></span></b></span></div>
<div style="background-color: white; color: #666666; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; line-height: 18.48px;">
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<b style="color: #666666;"><span style="font-size: large;"><br /><a href="http://www.venkatsacademy.com/2015/02/equations-of-motion-in-one-dimension.html" style="color: #888888; line-height: normal; text-decoration: none;"><span style="color: blue;">Equations of Motion in One Dimension</span></a><br /><span style="color: blue;"><br /></span></span></b><br />
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<b><span style="color: blue; line-height: normal;"><a href="http://www.venkatsacademy.com/2015/02/average-speed-average-velocity-and-acceleration.html" style="color: #888888; line-height: normal; text-decoration: none;"><span style="font-size: large;">Average Speed Average Velocity and Acceleration</span></a></span></b></div>
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<div style="line-height: 18.48px; text-align: start;">
<b><span style="color: blue;"><span style="font-size: large;"><a href="http://equations%20of%20motion%20in%20one%20dimension/" style="color: #888888; text-decoration: none;">Equation of motion in one dimensional motion</a> </span></span></b></div>
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<b><span style="color: blue;"><span style="font-size: large;"><br /></span></span></b></div>
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<h3 class="post-title entry-title" itemprop="name" style="font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
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<span style="color: #33aaff;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solution-eleven.html" style="color: #33aaff; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Eleven</span></a></span></span></h3>
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<span style="line-height: normal;">
<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-twelve.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Twelve</span></a></span></span></h3>
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<span style="line-height: normal;">
<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-thriteen.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Thirteen</span></a></span></span></h3>
<h3 class="post-title entry-title" itemprop="name" style="font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<span style="line-height: normal;">
<span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-and-two-dimensions-problems-with-solutions-thirteen.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One and two Dimensions Problems with Solutions Fourteen</span></a></span></span></h3>
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<a href="http://www.venkatsacademy.com/2017/02/vector-resolution-and-laws-of-vectors-video-lesson.html" style="color: #888888; text-decoration: none;">Vector resolution and Laws of Vectors Video lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-51895919891171876512017-02-04T18:22:00.001-08:002017-02-04T18:22:04.099-08:00Equations of Motion along straight line and Freely Falling Video lesson<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Equations of motion in one
dimensional motion<o:p></o:p></b></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">We know that a body in one
dimensional motion has displacement, velocity which is different at initial
level and final level and hence the body is also having some acceleration. The
body is covering some displacement in a specified time. Taking this into
consideration, we can obtain the relation between the above mentioned physical
quantities using the equations of motion. They relate some of the above
mentioned physical quantities and the relation is among the four quantities. If
we know any of the three, by using appropriate equation, we can find the
unknown physical quantity using the relations available to us.<o:p></o:p></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<center>
<iframe allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/VtrdguW_zMI" width="560"></iframe></center>
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b><br /></b></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Equation of motion for a freely
falling body<o:p></o:p></b></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">A body starting from the state of
rest from a certain height and falling vertically downwards under the influence
of gravitational force is called as a freely falling body. For a freely falling
body, initial velocity is zero and the displacement of the body is nothing but
the height of fall in a given time. As the body falls down, its velocity
increases and hence it is under acceleration and it is due to gravitational
force. This acceleration due to the earth is called acceleration due to gravity
and it is constant at a given place. Taking this into consideration, we can
rewrite the equations of motion as shown in the video lecture below. The time
taken by the body to reach the ground from the maximum height is called time of
descent.<o:p></o:p></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<br />
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<center>
<iframe allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/2mbcoAmxb7M" width="560"></iframe></center>
<br />
<div style="background-color: white; color: #666666; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; line-height: 18.48px;">
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<div style="background-color: white; color: #666666; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; line-height: 18.48px;">
<span style="color: blue;"><b><span style="font-size: large;"><span style="line-height: 18.48px;"><a href="http://venkatsacademy.blogspot.in/2014/11/distance-and-displacement-comparison.html" style="color: #888888; text-decoration: none;">Distance and displacement </a></span><span style="background-color: transparent; line-height: 18.48px;">comparison</span></span></b></span></div>
<div style="background-color: white; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; line-height: 18.48px;">
<b style="color: #666666;"><span style="font-size: large;"><br /><a href="http://www.venkatsacademy.com/2015/02/equations-of-motion-in-one-dimension.html" style="color: #888888; line-height: normal; text-decoration: none;"><span style="color: blue;">Equations of Motion in One Dimension</span></a><br /><span style="color: blue;"><br /></span></span></b><div style="color: #666666;">
<b><span style="color: blue; line-height: normal; text-align: start; text-decoration: none;"><a href="http://www.venkatsacademy.com/2015/02/average-speed-average-velocity-and-acceleration.html" style="color: #888888; line-height: normal; text-align: start; text-decoration: none;"><span style="font-size: large;">Average Speed Average Velocity and Acceleration</span></a></span></b></div>
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<b><span style="color: blue; font-size: large;"><br style="text-align: start;" /></span></b><div style="color: #666666; line-height: 18.48px; text-align: start;">
<b><span style="color: blue;"><span style="font-size: large;"><a href="http://equations%20of%20motion%20in%20one%20dimension/" style="color: #888888; text-decoration: none;">Equation of motion in one dimensional motion</a> </span></span></b></div>
<div style="color: #666666; line-height: 18.48px; text-align: start;">
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<b style="line-height: normal;"><span style="color: blue;"><span style="font-size: large;"><a href="http://www.venkatsacademy.com/2015/03/problems-on-motion-of-body-along-straight-line.html" style="color: #888888; text-decoration: none;">Problems on Motion of a Body Along a Straight Line</a> </span></span></b></div>
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<span style="line-height: normal;"><b><span style="color: blue;"><span style="font-size: large;"><a href="http://www.venkatsacademy.com/2015/07/acceleration-due-to-gravity-and-one.html" style="color: #888888; text-decoration: none;">Acceleration due to gravity and One Dimensional Motion Equations</a> </span></span></b></span></div>
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<div style="line-height: 18.48px;">
<b style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-one.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions One</span></a></b></div>
<div style="line-height: 18.48px;">
<h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-and-two-dimensions-problems-solutions-eight.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension and Two Dimension Problems with Solutions Eight</span></a></span></h3>
<h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-and-problems-and-solutions-nine.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Nine</span></a></span></h3>
<h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-ten.html" style="color: #33aaff; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Ten</span></a></h3>
<div>
<h3 class="post-title entry-title" itemprop="name" style="font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<span style="color: #33aaff;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solution-eleven.html" style="color: #33aaff; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Eleven</span></a></span></h3>
<h3 class="post-title entry-title" itemprop="name" style="font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-twelve.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Twelve</span></a></span></h3>
<h3 class="post-title entry-title" itemprop="name" style="font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-thriteen.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Motion in One Dimension Problems with Solutions Thirteen</span></a></span></h3>
</div>
<div>
<br /></div>
</div>
</div>
</div>
</span></div>
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</div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-37347691842583741492017-02-04T18:15:00.001-08:002017-02-04T18:15:10.631-08:00Motion in a Straight line Introduction and Average Velocity Video Lesson<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Motion in a straight line an
introduction<o:p></o:p></b></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">Studying the motion of the body without
bothering about the forces acting on it is done in kinematics. We treat body as
a combination of identical point sized objects and they have negligible dimensions.
All laws of mechanics were in principle discussed with the point sized
particles and as the body is the combination of similar particles, under ideal
conditions the laws are applicable to bodies also. Here we are dealing with
bodies moving with a velocity much lesser than the velocity of the light. In
this particular case, body is moving only along one dimension either along X,Y
or Z axis. This is called one dimensional motion and it is changing its
position with respect to time and surroundings.</span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">To measure the change of the
position, we have terms like distance, speed. Distance is the actual path
traveled by a body and the speed is the rate of change of distance with respect
to time. Both distance and speed are treated as scalars and they can be
understood by stating their magnitude alone and they don’t need direction.<o:p></o:p></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">Displacement is the shortest
distance between initial and final positions in specified direction and it is
treated as vector quantity. They can be understood completely only when both
magnitude and directions are given to us. Velocity is defined as the rate of
change of displacement and it is also a vector quantity.<o:p></o:p></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br />
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<center>
<iframe allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/KPSzRI5a-o8" width="560"></iframe></center>
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b><br /></b></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Average velocity<o:p></o:p></b></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;">If a particle is not changing its
velocity with respect to time, then it is said to be in uniform velocity. In
this case at any given interval of time, the particle will have same constant
velocity and it is same every where. But it is not same every where. If a body
is changing its velocity with respect to time, then it is having acceleration
and we would like to measure the average velocity in the given case. Average
velocity is defined as the ratio of total displacement covered by a body in the
total time. Taking this concept into consideration, we can find average
velocity when time is shared and displacement is shared as shown in the video
below.<o:p></o:p></span></span></div>
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<b><span style="color: blue;"><span style="font-size: large;"><a href="http://equations%20of%20motion%20in%20one%20dimension/" style="color: #888888; text-decoration: none;">Equation of motion in one dimensional motion</a> </span></span></b></div>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-and-solutions-six.html" style="color: #888888; text-decoration: none;">Motion in One Dimension Problems with Solutions Six</a></span></h3>
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<span style="color: #33aaff;"><a href="http://www.venkatsacademy.com/2016/10/motion-in-one-dimension-problems-with-solutions-seven.html" style="color: #33aaff; text-decoration: none;">Motion in One Dimension Problems with Solutions Seven</a></span></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-89463195298404695532017-02-04T05:14:00.003-08:002017-02-04T05:14:30.976-08:00Uses and Applications of Dimensional Analysis Video Lesson<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">Dimensional formula is a
representation of a physical quantity in terms of fundamental quantities. We
represent mass with M, length with L and time with T in dimensional formula.
Dimensional analysis can be done basing on principle of homogeneity and
according to it the dimensions of left hand side and right hand side of the
equation has to be equal. It mean to tell us that we can add or subtract only
similar physical quantities but not dissimilar ones. Using the dimensional
formula and analysis, we can convert a physical quantity from one system of
unit to other. </span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Conversion of physical Quantity from one system to other</b></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">For example we can that the energy is measured in the unit joule
in SI system and erg in CGS system. Because they are the units of same physical
quantity in different system of units, they shall be having some relation
between them. We can find that relation using the principle of homogeneity.<o:p></o:p></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Checking the correctness of given equation</b></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="font-size: large;">We can also use the dimensional
analysis to check the correctness of a given equation. We have so many
equations in physics and before they are correct as per the subject, they have
to be correct conceptually. That can be verified using the dimensional
analysis. Here we are depending on the concept of principle of homogeneity and
according to it LHS and RHS of a equation shall have same dimensions of the
physical quantities. If the sum of left hand side physical quantities is
velocity then the right hand side sum or difference of the terms shall also be
the velocity. In the other sense, we are adding or subtracting physical quantities
of the same nature but not different.<o:p></o:p></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
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<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Finding Relation among Physical Quality</b></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="font-size: large;">We can also use dimensional analysis
to find the relation between physical quantities basing on principle of
homogeneity. We will simple equate the dimensions of left hand side of the
equation with the right hand side equation so that we will be getting
mathematical equations and by solving them, we can get the dimensions of the
physical quantities. Here we will be able to find only the relation but we can
not find the proportional constant values using this method. This is one
limitation of the dimensional analysis.<o:p></o:p></span></span></div>
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<b><span style="font-size: large;"><span style="font-family: "times new roman"; line-height: 18.48px;"><span style="font-family: "trebuchet ms", trebuchet, verdana, sans-serif; line-height: 27.72px;"><span style="line-height: 18.48px;"><a href="http://venkatsacademy.blogspot.com/2014/09/dimensional-formula-for-physical.html" style="color: #888888; text-decoration: none;" target="_blank">Writing dimensional formula</a></span></span></span><br /><span style="font-family: "times new roman"; line-height: 18.48px;"><span style="font-family: "trebuchet ms", trebuchet, verdana, sans-serif; line-height: 27.72px;"><br /><span style="color: #888888; line-height: 18.48px;"><a href="http://venkatsacademy.blogspot.com/2014/09/errors-and-approximations-in-physical.html" style="color: #888888; line-height: 18.48px; text-decoration: none;" target="_blank">Errors and approximations</a></span></span></span><br /><br /><span style="color: #888888; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-and-solutions-one.html" style="color: #888888; text-decoration: none;">Units and Dimensions Problems and Solutions One</a></span></span></b><h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-andsolutions-5.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions Five</span></a></span></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-54344321230755792232017-02-04T04:54:00.001-08:002017-02-04T05:15:29.582-08:00Principle of Homogeneity and Limitations of Dimensional Analysis Video Lesson<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">Dimensional formula is a
representation of a physical quantity in terms of fundamental quantities. When
we write a physics equation, it shall be in such a way that the dimensions of
left hand side of the equation has to be equal to the dimensions of all
physical quantities along the right hand side of the equation. This is possible
only when you add or subtract similar quantities at either left or right side
of the equation. </span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">It is simple understanding that velocity can be added only
with velocity to get another velocity We can not add velocity with force and
the summation can not give either velocity or force and the summation is
meaning less.This principle is called principle of homogeneity. If any equation
is not satisfying the law, then it cannot be correct with respect to physics.
But we need to be careful that the dimensionally correct equation can not be
correct with respect to physics. Being satisfying the principle of homogeneity
is the fundamental condition to be accepted as a physics equation. <o:p></o:p></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
</div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Limitations of Dimensional Analysis</b></span></span><br />
<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="font-size: large;">Dimensional formulas helps us to
understand the relation among the physical quantities and it helps us also in
converting the physical quantity from one system of unit to other system of
unit. But they have certain limitations. Trigonometric and exponential
functions won’t have any dimensions and if they are involved in any equation,
we can not solve them basing on dimensional analysis. If the equation on the
right hand side is depending on more fundamental quantities than the left hand
side, then we cannot solve the equation basing on dimensional analysis. We also
won’t be able to find out the proportionality constants of a science equation
using the dimensional analysis.<o:p></o:p></span></span></div>
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<b><span style="font-size: large;"><span style="font-family: "times new roman"; line-height: 18.48px;"><span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; line-height: 27.72px;"><span style="line-height: 18.48px;"><span style="color: blue;"><span style="color: #888888;"><a href="http://venkatsacademy.blogspot.com/2014/09/measuring-physical-quantites-with-units.html" style="color: #888888; text-decoration: none;" target="_blank">Units and measurement</a></span></span></span></span></span><br /><span style="font-family: "times new roman"; line-height: 18.48px;"><span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; line-height: 27.72px;"><span style="line-height: 18.48px;"><span style="color: blue;"><br /></span></span></span></span></span></b></div>
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<b><span style="font-size: large;"><span style="font-family: "times new roman"; line-height: 18.48px;"><span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; line-height: 27.72px;"><span style="line-height: 18.48px;"><a href="http://venkatsacademy.blogspot.com/2014/09/dimensional-formula-for-physical.html" style="color: #888888; text-decoration: none;" target="_blank">Writing dimensional formula</a></span></span></span><br /><span style="font-family: "times new roman"; line-height: 18.48px;"><span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; line-height: 27.72px;"><br /><span style="color: #888888; line-height: 18.48px;"><a href="http://venkatsacademy.blogspot.com/2014/09/errors-and-approximations-in-physical.html" style="color: #888888; line-height: 18.48px; text-decoration: none;" target="_blank">Errors and approximations</a></span></span></span><br /><br /><span style="color: #888888; font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-and-solutions-one.html" style="color: #888888; text-decoration: none;">Units and Dimensions Problems and Solutions One</a></span></span></b><br />
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-and.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions Two</span></a></span></h3>
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<a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-andsolutions-3.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions Three</span></a></h3>
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<a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-andsolutions-4.html" style="color: #888888; text-align: justify; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions Four</span></a></div>
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<span style="color: #888888;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-andsolutions-5.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions Five</span></a></span></h3>
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<span style="color: #888888; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-andsolutions-6.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions Six</span></a></span></h3>
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2017/02/units-and-dimensions-introduction-video-lesson.html" style="color: #888888; text-decoration: none;">Units and Dimensions an Introduction Video Lesson</a></span></h3>
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<a href="http://www.venkatsacademy.com/2017/02/uses-and-applications-of-dimensional-analysis-video-lesson.html" style="color: #888888; text-decoration: none;">Uses and Applications of Dimensional Analysis Video Lesson</a></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-86938432322419711962017-02-03T17:17:00.000-08:002017-02-04T04:55:22.548-08:00Units and Dimensions an Introduction Video Lesson<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN"><span style="font-size: large;">We are dealing with the basics of concepts units and dimensions in a video lesson. Physical quantity is a way of understanding nature and its physics applications and to measure them, we use units.</span></span></div>
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<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
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<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Introduction to Units<o:p></o:p></b></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;">Physical quantities are helpful in
understanding the nature in terms of measurement. We try to understand the
world around us with minimum physical quantities and they are the alphabets and
words of the physics language. If a physical quantity is independent of any
other physical quantities, they are called fundamental physical quantities and
they are the irreducible set of quantities to represent other things in the
nature around. Length, mass and time are basic examples of fundamental physical
quantities.</span></span></div>
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<span lang="EN-IN"><span style="font-size: large;"><br /></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;">Other physical quantities are derived basing on this fundamental
quantities and they are called derived physical quantities. Velocity,
acceleration and force are the some of the examples of derived physical
quantities. To measure this physical quantities, we need to use units. Unit is
a way of measuring the physical quantities in a standard way that is acceptable
to all.<o:p></o:p></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;">As physical quantities are of two
types, the units are also two types and they are fundamental and derived units.
To measure the fundamental physical quantities, we have fundamental units with
different kinds of systems like FPS,CGS,MKS and SI system of units. We can
further define and write units for the derived quantities based on the the
fundamental system of units. At present we are using SI system as the standard
system to measure the fundamental physical quantities. <o:p></o:p></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;"><br /></span></span>
<span lang="EN-IN"><span style="color: blue; font-size: large;"><b>Introduction to Dimensions<o:p></o:p></b></span></span></div>
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<span lang="EN-IN"><span style="font-size: large;">Each physical quantity is either a
fundamental or derived physical quantity. Derived physical quantity is obtained
from fundamental quantity by manipulating the fundamental quantities as per the
requirement of the concepts and applications. The representation of a physical
quantity in terms of fundamental quantities is called dimensional formula and
the powers to which the fundamental quantities are raised are called
dimensions. Thus using the dimensional formula, we will be knowing the content
of fundamental quantities in it.<o:p></o:p></span></span></div>
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<span style="font-family: "times new roman"; line-height: 18.48px; text-align: justify;"><span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; line-height: 27.72px;"><span style="line-height: 18.48px;"><span style="color: blue;"><span style="color: #888888; font-size: large;"><a href="http://venkatsacademy.blogspot.com/2014/09/measuring-physical-quantites-with-units.html" style="color: #888888; text-decoration: none;" target="_blank"><b>Units and measurement</b></a></span></span></span></span></span><br />
<span style="font-family: "times new roman"; line-height: 18.48px; text-align: justify;"><span style="font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; line-height: 27.72px;"><span style="line-height: 18.48px;"><span style="color: blue; font-size: large;"><b><br /></b></span></span></span></span></div>
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<span style="color: #888888; font-family: "trebuchet ms" , "trebuchet" , "verdana" , sans-serif; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-and-solutions-one.html" style="color: #888888; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; text-decoration: none;"><span style="font-size: large;">Units and Dimensions Problems and Solutions One</span></a></span><br />
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<span style="color: #888888; font-size: large; text-decoration: none;"><a href="http://www.venkatsacademy.com/2016/10/units-and-dimensions-problems-andsolutions-6.html" style="color: #888888; text-decoration: none;">Units and Dimensions Problems and Solutions Six</a></span></h3>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-91052561110293310662017-01-17T06:05:00.002-08:002017-02-03T17:22:16.364-08:00Oscillations Problems with Solution Three<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
are solving series of problems based on the concept of oscillation and simple
harmonic motion. If the acceleration of a body is directly proportional to the
displacement and opposite to it, then the motion is said to be simple harmonic
motion. This can be done by many type of systems and if they are obeying the
above mentioned condition, we can get the time period or frequency and it is
constant all over the oscillatory motion. A loaded spring do oscillate when it
is slightly disturbed. In practical way, any body won’t continue its
oscillatory motion for ever and and due to air resistance, it slowly decreases
and finally comes to the state of the rest. This kind of motion is called
damped oscillatory motion and it that has to be continued, it shall have some
external force support and that kind of motion is called forced oscillation.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><br /></span></div>
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<span style="font-size: 14pt;">We
need to find the time period of pendulum of infinite length and the problem is
as shown in the diagram below.</span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
need to write the equation for the torque as the product of force and
perpendicular distance. Force is nothing but the weight of the body and
distance is found as shown in the diagram below. We need to write further
torque as the product of moment of inertia and angular acceleration basing on
its definition. Thus we can equation for the acceleration and hence the time
period as shown in the diagram below.<o:p></o:p></span></div>
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<span style="font-size: 14pt;"><b><span style="color: blue;">Problem</span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Frequency
of a particle in SHM is given to us as shown below. We need to find the maximum
speed that the particle can reach in the oscillatory motion.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://4.bp.blogspot.com/-VMs0p04yfak/WH4jvEsSRHI/AAAAAAAAIlw/yzvvGc0ezU0v-7OlqtCLOUIVt4DS5772wCLcB/s1600/32.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="466" src="https://4.bp.blogspot.com/-VMs0p04yfak/WH4jvEsSRHI/AAAAAAAAIlw/yzvvGc0ezU0v-7OlqtCLOUIVt4DS5772wCLcB/s640/32.jpg" width="640" /></a></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">The
restoring force acting on the spring is nothing but the weight of the body and
hence we can find the maximum possible displacement by equating them. By
comparing that with the standard equation, we can find angular velocity and
hence the maximum speed of the particle as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem</span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
particle of known mass is attached to three springs as shown in the diagram
below. If the particle is slightly disturbed, we need to know the time period
of the system.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span style="font-size: 14pt;"> </span><a href="https://1.bp.blogspot.com/-4j0JPvRCe-I/WH4kIQBiiLI/AAAAAAAAIl8/GhRUGwWfK5QrDFH5FICjVUvZ3xCuKFhMACLcB/s1600/33.jpg" imageanchor="1" style="font-size: 14pt; margin-left: 1em; margin-right: 1em; text-align: center;"><img border="0" height="640" src="https://1.bp.blogspot.com/-4j0JPvRCe-I/WH4kIQBiiLI/AAAAAAAAIl8/GhRUGwWfK5QrDFH5FICjVUvZ3xCuKFhMACLcB/s640/33.jpg" width="466" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">The
resultant force acting on the mass using the vector laws of addition. We need
to resolve the force into components and add as shown in the diagram below. By
equating it to the restoring force, we can write the equation for the time
period as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Two
blocks are kept one over the other and the lower surface is smooth and the
connecting surface is rough as shown in the diagram below. The system is
connected to a rigid support with a spring and the time period of the system is
given to us in the problem. We need to find the mass of the upper block and the
coefficient of friction so that there is no slipping between the two blocks.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://4.bp.blogspot.com/-ziltqo0Qxbc/WH4kPwveeLI/AAAAAAAAImA/isGFMYc6ZVsnvFsZfhlj9pgCuQVJNZx6ACLcB/s1600/36.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://4.bp.blogspot.com/-ziltqo0Qxbc/WH4kPwveeLI/AAAAAAAAImA/isGFMYc6ZVsnvFsZfhlj9pgCuQVJNZx6ACLcB/s640/36.jpg" width="562" /></a></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
know the equation for the time period of a loaded spring and using that data,
we can find the mass of the upper body as shown in the diagram below. Further
equating the frictional force to the restoring force, we can solve the problem
as shown in the diagram below.<o:p></o:p></span></div>
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<a href="http://www.venkatsacademy.com/2017/01/oscillations-problems-with-solution-one.html" style="color: #888888; text-decoration: none;"><span style="font-size: large;">Oscillations Problems with Solution One</span></a></b></b></b></b></b></b></b></h3>
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<h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative; text-align: start;">
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<span style="color: #888888; font-size: large; text-decoration: none;">Oscillations Problems with Solution Two</span></a></b></b></h3>
<h3 class="post-title entry-title" itemprop="name" style="font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative; text-align: start;">
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<b style="font-family: "times new roman"; line-height: 18.48px; text-align: justify;"><b style="font-family: "trebuchet ms", trebuchet, verdana, sans-serif; line-height: 27.72px;">
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-2541141348515435252017-01-17T05:55:00.001-08:002017-01-17T06:06:08.157-08:00Oscillations Problems with Solution Two<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We are solving series of problems based on the concept of oscillations. It is also called vibratory or harmonic motion where there is to and fro motion that gets repeated at regular intervals of time. The vertical projection of uniform circular motion is simple harmonic and we have derived equations for displacement, velocity and acceleration for the body in simple harmonic motion based on that. The time taken to complete one oscillation is called time period and the number of oscillations per one second is called frequency.</span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><br /></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem</span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><br /></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
body of known mass is connected with two springs and they in tern are connected
to rigid support as shown in the diagram below. We need to find the effective
time period of the system and the problem is as shown in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">When
ever the body is slightly disturbed, it starts oscillating and one spring
expands and the other spring contracts by the same magnitude. As the force
acting on both of them is same, the two springs behaves as if like they are
connected in series and we can find the effective spring constant of the system
and time period as shown in the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
spring of spring constant K and length L is cut into two parts and the relation
between the lengths of two parts and their ratio is given to us in the problem
as shown in the diagram below. We need to measure the spring constant of one
part of the spring.<o:p></o:p></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Spring
constant is the measure of nature of spring and it depends on the length of spring
in the inverse proportional ratio. Taking that into consideration, we need to
solve the problem as shown in the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
piece of wood known dimensions is given to us and its density is also known to
us as given in the problem below. It is floating in water with one surface
vertical to the surface. It is pushed down and released and it starts
oscillating. We need to measure the time period of the system.<o:p></o:p></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">When
ever a force is applied on the body, there is buoyant force also and the system
starts executing oscillatory motion with a certain restoring force. By
comparing that with the standard equation, we can solve the problem as shown in
the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
cylindrical piston is used to close a cylinder with a certain gas and when it
is slightly distributed, it starts oscillating in simple harmonic motion. We
need to find the time period of the system and the problem is as shown in the
diagram below.<o:p></o:p></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution</span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><br /></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">By
comparing it with the standard equation, we can solve the problem as shown in
the diagram below. We need to identity the restoring force and the equation for
the acceleration of the body in SHM. The problem is solved as shown in the
diagram below. Here in the first case, we get the equation for the force.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
need to equate to the product of mass and acceleration and we further need to
substitute the value of acceleration for a body in SHM so that we can compare
it with the standard equation. Thus we can get the time period of the system as
shown in the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
spherical ball of known mass and radius is rolling with out slipping on a
concave surface of known radius as shown in the diagram below. If the
oscillations are small, we need to find the time period of the system.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Let
us consider the spherical ball is at a particular position and we can find the
angle basing on the definition as shown in the diagram below. As we know the
equation for the acceleration of a body sliding on a inclined surface, by
writing that equation, we can get that and find the time period as shown in the
diagram below.<o:p></o:p></span></div>
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<br />
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<b><span style="line-height: normal;">Time Period of Simple pendulum </span><br /><span style="line-height: normal;"><br /></span></b></div>
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<span style="color: #888888; line-height: normal;"><b><br /></b></span><span style="color: #888888; line-height: normal;"><b><a href="http://venkatsacademy.blogspot.in/2014/12/loaded-spring-in-simple-harmonic-motion.html" style="color: #888888; text-decoration: none;">Loaded spring in simple harmonic motion</a> </b></span></div>
<div style="line-height: 18.48px;">
<span style="color: #888888; line-height: normal;"><b><br /></b></span></div>
<div style="line-height: 18.48px;">
<b style="color: #888888; line-height: normal;"><a href="http://venkatsacademy.blogspot.in/2014/12/damped-and-forced-oscillations.html" style="color: #888888; line-height: normal; text-decoration: none;">Damped Oscillations and Forced Oscillations</a></b></div>
</div>
</div>
</b></b></div>
</span></b></b></div>
Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-62174170827340753.post-37236102668735183412017-01-17T04:23:00.001-08:002017-01-17T04:23:22.201-08:00Oscillations Problems with Solution One <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
are solving series of problems based on the concept of oscillations.
Oscillation is a kind of motion where the body oscillates about a fixed point
called mean position and all oscillatory motions are periodic. It means
oscillatory motion is repeated at regular intervals of time. We need to
understand that all oscillatory motions are periodic but all periodic motions
are not oscillatory. If oscillatory motion is also satisfying a condition like
displacement is directly proportional to acceleration and acceleration is
always directed to wards the mean position, we call that kind of oscillatory
motion as simple harmonic motion. Simple pendulum is one example that executes
simple harmonic motion when it is sightly disturbed from its mean position. We
have derived equation for displacement, velocity and acceleration for a body in
simple harmonic motion.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">To
a body in a simple harmonic motion, velocity is represented as shown in the
equation below. We need to measure maximum acceleration that the body can get
in the given conditions.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><br /></span></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://4.bp.blogspot.com/-Uz-xEWW_H0M/WH4LW_prXgI/AAAAAAAAIkE/esKMdSXnA7kG3anCRDRKl4og4Om19eQUQCLcB/s1600/1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="270" src="https://4.bp.blogspot.com/-Uz-xEWW_H0M/WH4LW_prXgI/AAAAAAAAIkE/esKMdSXnA7kG3anCRDRKl4og4Om19eQUQCLcB/s640/1.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
have all ready derived equation for the velocity of the particle in simple
harmonic motion. We need to get the given equation in the terms of the standard
equation and the problem can be solved as shown in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><br /></span></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-XB2RjPYsiaY/WH4LW4hg6gI/AAAAAAAAIj8/xPmAy54vaoU1ODkZRCcEeiEDR2K6eQo0ACLcB/s1600/1s.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="396" src="https://1.bp.blogspot.com/-XB2RjPYsiaY/WH4LW4hg6gI/AAAAAAAAIj8/xPmAy54vaoU1ODkZRCcEeiEDR2K6eQo0ACLcB/s640/1s.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem
<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
particle starts from mean position to a new position and it is as shown in the
diagram below. Its amplitude and time period is given to us in the problem. We
need to find the displacement where the velocity is half of the maximum
velocity.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://4.bp.blogspot.com/-V8vp9dhN-XM/WH4LW-PG-UI/AAAAAAAAIkA/mikPGevDo9ALd2NlGrZGN3SECzLL_8NLQCLcB/s1600/2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="474" src="https://4.bp.blogspot.com/-V8vp9dhN-XM/WH4LW-PG-UI/AAAAAAAAIkA/mikPGevDo9ALd2NlGrZGN3SECzLL_8NLQCLcB/s640/2.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
know that the particle in simple harmonic motion has maximum velocity at the
mean position. As per the given problem at a given instant, velocity of the
particle is half of that maximum. Taking that into consideration and
substituting the data in the standard format, we can solve the problem as shown
in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://2.bp.blogspot.com/-2oZflESSfdk/WH4LXXtuILI/AAAAAAAAIkI/afkr0ZzVucILm_6VQD41lUVuzbY04gRpQCLcB/s1600/2s.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://2.bp.blogspot.com/-2oZflESSfdk/WH4LXXtuILI/AAAAAAAAIkI/afkr0ZzVucILm_6VQD41lUVuzbY04gRpQCLcB/s640/2s.jpg" width="592" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Two
different particles are in simple harmonic motion and their displacements are represented as the given equations of the problem. We
need to find the resultant amplitude of the combination. Problem is as shown
below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-akjqPxDt5Rs/WH4LXXh685I/AAAAAAAAIkM/pe3nV8eUk64KhFJO1jhnC7mfziJ9Sl_BQCLcB/s1600/3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="200" src="https://1.bp.blogspot.com/-akjqPxDt5Rs/WH4LXXh685I/AAAAAAAAIkM/pe3nV8eUk64KhFJO1jhnC7mfziJ9Sl_BQCLcB/s640/3.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">When
we add to oscillatory motion, we need to get a oscillatory motion. The
resultant amplitude can be found using the vector addition equation and the
solution is as shown in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-cwyYrbeOhIk/WH4LXYlLq9I/AAAAAAAAIkQ/uUHER2r_igQc2HThVJiPHRDCyHfCNRaNQCLcB/s1600/3s.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="490" src="https://1.bp.blogspot.com/-cwyYrbeOhIk/WH4LXYlLq9I/AAAAAAAAIkQ/uUHER2r_igQc2HThVJiPHRDCyHfCNRaNQCLcB/s640/3s.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
simple harmonic oscillator starts from extreme position and covers a half the
displacement in a given time. We need to measure the further time it is going
to take to reach the mean position and the problem is as shown in the diagram
below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://3.bp.blogspot.com/-KqpHqah1bec/WH4LXymINOI/AAAAAAAAIkU/NnvTr86S2BAtUaXvwGWu5ek2VIsCnoEuwCLcB/s1600/4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="310" src="https://3.bp.blogspot.com/-KqpHqah1bec/WH4LXymINOI/AAAAAAAAIkU/NnvTr86S2BAtUaXvwGWu5ek2VIsCnoEuwCLcB/s640/4.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">As
the particle is here starting from the mean position, we need to know that it
has some initial phase that is ninety degree. We know that the particle takes
one forth of the time period to reach from extreme to mean position and to
measure the remaining time to cover half amplitude to, we need to subtract from
it as shown in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-ZIIPbFB8x7s/WH4LX9mzE2I/AAAAAAAAIkY/5D2mHWZzoa0KMCylFp0dQNSzNOjmHKkagCLcB/s1600/4s.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://1.bp.blogspot.com/-ZIIPbFB8x7s/WH4LX9mzE2I/AAAAAAAAIkY/5D2mHWZzoa0KMCylFp0dQNSzNOjmHKkagCLcB/s640/4s.jpg" width="488" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Number
of springs are connected in series as shown in the problem to a a given mass
and the system is allowed to oscillate. We need to measure the time period of
oscillation of that system.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://4.bp.blogspot.com/-4TnSpbphVdM/WH4LYFM3wDI/AAAAAAAAIkc/0CVZDCa9Mks-ki0Oa1vKErNa_ZvEezVXQCLcB/s1600/6.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="332" src="https://4.bp.blogspot.com/-4TnSpbphVdM/WH4LYFM3wDI/AAAAAAAAIkc/0CVZDCa9Mks-ki0Oa1vKErNa_ZvEezVXQCLcB/s640/6.jpg" width="640" /></a></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
know that when the springs are in series, the force acting on all of them is
same and the extension in the spring is different and it depends on the nature
of the spring. Using the common formula for the time period of the system and
further simplify the problem as shown in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<br />
<div class="separator" style="clear: both; text-align: center;">
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<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
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</span></b></b></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-63989315173940567852017-01-12T08:19:00.003-08:002017-01-17T04:24:22.036-08:00Thermodynamics Problems with Solutions Five<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<b><span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
are </span></b><span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">solving series of problems based on the concepts of
thermodynamics. We also deal about calorimetry in this chapter which deals
about conversion of heat energy to other forms of energies and its applications
further. We do define terms like specific heat and latent heat to explain this
properties and they are the basic terms of heat concepts. When there is a
change in the temperature, we need to deal with specific heat concept and when
there is a change of state, we need to study it in terms of latent heat and
during this process, all the supplied heat energy is used to change the state
of the system and hence its temperature remains constant.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">During
an adiabatic process,pressure of the gas is proportional to the cube of the
temperature and basing on that we need to find the ratio of specific heats of
the gas. Problem is as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
need to take the relation between pressure and temperature and taking that into
consideration with the given data, we can get the relation between pressure and
temperature in the adiabatic process and the problem can be solved as shown in
the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
metal sphere of known radius and specific heat is given to us and it is
rotating about its own axis with certain rotations per second. When it is
stopped half of its energy is converted into heat and we need to measure the
raise in the temperature of the system and the problem is as shown in the
diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://2.bp.blogspot.com/-sppKv9PeyB4/WHerx6rxovI/AAAAAAAAIjE/hle9H5r9cWM3uJkO9_lx2e8rO6LH0MTfwCLcB/s1600/38.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="460" src="https://2.bp.blogspot.com/-sppKv9PeyB4/WHerx6rxovI/AAAAAAAAIjE/hle9H5r9cWM3uJkO9_lx2e8rO6LH0MTfwCLcB/s640/38.jpg" width="640" /></a></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">As
the body is rotating it has rotational kinetic energy and half of it is
converted into heat energy as per the given problem. Taking law of conservation
of the energy, we can solve the problem as shown in the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">The
relation between internal energy,pressure and volume is given to us as shown in
the diagram below. We need to find the ratio of specific heats and some
constants are also available in the problem.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-EBWjImqyvVg/WHesAeG_ryI/AAAAAAAAIjM/MMZfvFk98Dsm25QV_skK3qiBmAs1CQPbgCLcB/s1600/39.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="364" src="https://1.bp.blogspot.com/-EBWjImqyvVg/WHesAeG_ryI/AAAAAAAAIjM/MMZfvFk98Dsm25QV_skK3qiBmAs1CQPbgCLcB/s640/39.jpg" width="640" /></a></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
need to differentiate the given equation to get the change in internal energy
and hence it can be expressed in terms of specific heat of the gas at constant
volume. Problem can be further solved as shown in the diagram below.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Work
done by a system under isothermal conditions has to be determined that change
its volume from one to other and satisfy the given equation. Problem is as
shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Solution<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Relation
of pressure with other physical quantities is given to us as shown in the
diagram below. We need to measure the work done and the pressure is not
constant here. So to get the work done we shall integrate the pressure with the
change in the volume as shown in the diagram below. By simplifying the equation
further, we can get the solution as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-11309866132819456622017-01-12T08:12:00.000-08:002017-01-12T08:20:10.104-08:00Thermodynamics Problems with Solutions Four<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
are solving series of problems based on the concept of thermodynamics. It is a
branch of physics that deals with the heat energy conversion into other forms
and its applications. Carnot engine is a ideal heat engine that converts the
given heat into work under a cyclic process. As the process is cyclic process,
internal energy remains same and all the supplied heat is going to be converted
into work. But we know that heat is a disordered format of energy and it cannot
be completely converted into work. That is the reason why, heat engine cannot
have hundred percent efficiency. Heat engine will have three basic parts by
name source, working substance and sink. Source do supply heat, working
substance do convert the heat into work and the extra generated heat to the
sink.<o:p></o:p></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Pressure
and volume graph for two different gases during adiabatic process is given to
us as shown in the diagram below. We need to know which graph belongs to which
gas.<o:p></o:p></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">From
the pressure and volume graph, it can be found that the slope of the graph is
directly proportional to the ratio of specific heats of the gas. It is clear
from the graph that the slope of the second curve is more and hence it shall be
the gas with more ratio of specific heat value. Solution of the problem is as
shown in the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">A
refrigerator is placed in a room of temperature of 300 kelvin and the system
temperature is 264 kelvin. We need to measure the how many calories of heat
shall be delivered to the room to the room for each kilo kelvin of energy
consumed by refrigerator by the system ideally. Problem is as shown in the diagram
below.<o:p></o:p></span></div>
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<br /></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
know that proficiency of a refrigerator is the amount of work done when
compared with the heat energy supplied to the system. It can also be expressed
in terms of temperature as shown in the diagram and is solved as shown in the
diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">In
carnot’s engine efficiency is 40 % for a certain temperature of the hot source.
To increase the efficiency of the system by 50 %, what shall be the source
temperature and the problem is as shown in the diagram below.<o:p></o:p></span></div>
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<div class="separator" style="clear: both; text-align: center;">
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
know that the efficiency of heat engine is the magnitude of the work done when
compared with the heat energy supplied and it can be expressed in terms of
absolute temperature as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
need to apply it for two cases and solve the problem as shown in the diagram
below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">The
relation between temperature and pressure is given to us for a certain gas and
we need to find the specific heats ratio of that gas. Problem is as shown in
the diagram below. <o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">This
process is adiabatic process and we need to write the given relation between
pressure and temperature from the given format to the standard format so that
we can solve and find the ratio of specific heats of the given gas as shown in
the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">When
a mono atomic gas expands at constant pressure, the percentage of the heat
supplied that increases the temperature of the gas and in doing external work
is how much is the problem and it is as shown in the diagram below.<o:p></o:p></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
know that if volume is kept constant, work done is zero and corresponding
specific heat is defined as per it and vice versa. When we write ratio of
specific heats, it becomes the ratio of heat supplied to the change in the
internal energy and internal energy change the temperature of the system. Thus
we can find the percentage value basing on the specific heats ratio value as
shown in the diagram below.<o:p></o:p></span></div>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-62174170827340753.post-86357705700935896532017-01-12T08:01:00.003-08:002017-01-12T08:12:49.968-08:00Thermodynamics Problems with Solutions Three<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<b><span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">We
</span></b><span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">are solving series of problem in the topic
thermodynamics. There are different types of process in thermodynamics.
Isobaric process means pressure of the system is kept constant. Isothermal
process means temperature of the system is kept constant. Adiabatic process
means heat energy of the system is kept constant. In each case, work done is
different and we need to measure as per the given process. Zeroth law of
thermodynamics is regarding temperature concept. First law of thermodynamics is
regarding conservation of energy and second law of thermodynamics is regarding
direction of flow of heat.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Three
samples of same gas has initially same volume. Volume of each one is doubled
and the process is different in each case. It is adiabatic, isobaric and
isothermal respectively . if all the final pressures are equal, we need to know
the initial pressures of the three and the problem is as shown in the diagram
below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://2.bp.blogspot.com/-9ufMFCnqMic/WHem_kDw84I/AAAAAAAAIho/DYM7EjoEuH8G2MtwBrkT0MzuHtmte5ZpwCLcB/s1600/18.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="434" src="https://2.bp.blogspot.com/-9ufMFCnqMic/WHem_kDw84I/AAAAAAAAIho/DYM7EjoEuH8G2MtwBrkT0MzuHtmte5ZpwCLcB/s640/18.jpg" width="640" /></a></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">The
relation between pressure and volume is different in each case. Isobaric
process means pressure remains constant and in that case, initial pressure is
equal to final pressure. In the case of isothermal process, as temperature is
constant, we can apply boyle’s law and find the final pressure in terms of
initial pressure. Adiabatic process get the ratio of specific heats into the
picture and taking that into consideration, we can solve the problem as shown
in the diagram below.<o:p></o:p></span></div>
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<br /></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">The
pressure inside a tyre at one temperature is given to us as four atmospheric
pressure and we need to find the temperature when the tyre bursts suddenly.
Problem is as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://4.bp.blogspot.com/-gPlNpFEN5w4/WHenPNCc4GI/AAAAAAAAIhw/lNwyh6I4D-85ghqcA0X9QamvXDlH8G6PQCLcB/s1600/19.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="272" src="https://4.bp.blogspot.com/-gPlNpFEN5w4/WHenPNCc4GI/AAAAAAAAIhw/lNwyh6I4D-85ghqcA0X9QamvXDlH8G6PQCLcB/s640/19.jpg" width="640" /></a></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">As
the tyre bursts suddenly, it is going to be a adiabatic process and the heat
energy of the system remains constant. We need to apply and find the relation
between temperature and pressure using the pressure and volume relation of the
adiabatic process and simplify the problem as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">There
are two cylinders with the same ideal gas at the same temperature. There are
pistons on both of them and their initial temperature is same. If one piston is
allowed to move freely and the other is fixed, we need to compare the
temperature of the second case when compared with the first case. Problem is as
shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">In
the first case pressure is kept constant and in the second case it is the
volume that is kept constant by sealing the system. Taking the respective
specific heats and using the equation for the heat energy in each case, we can
solve the problem as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">An
ideal gas after going through four thermodynamic states has come back to its
initial state. Heat energy in each case is given to us and work done in three
cases is given to us as shown in the diagram below. We need to find the work
done in the fourth case.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;">
<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">As
the process is cyclic, there is no change in the internal energy of the system
and as per the first law of thermodynamics, the heat energy supplied in this
case will be the work done itself. By equating the total heat energy with
proper sign to the total work done, we can solve the problem as shown in the
diagram below.<o:p></o:p></span></div>
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<br /></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Problem<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">PV
diagram of an ideal gas is as shown and we need to measure the work done during
a part as shown in the diagram below.<o:p></o:p></span></div>
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<br /></div>
<div class="separator" style="clear: both; text-align: center;">
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;"><b><span style="color: blue;">Solution<o:p></o:p></span></b></span></div>
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<br /></div>
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<span lang="EN-IN" style="font-size: 14.0pt; line-height: 115%; mso-ansi-language: EN-IN;">Work
is said to be done when there is a change in the volume of the system other
wise the work done is zero. In a PV diagram, work done is the area under the
graph. We can use the shape and find the area of the graph to solve the
problem.<o:p></o:p></span></div>
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<br /></div>
<br />
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<h3 class="post-title entry-title" itemprop="name" style="background-color: white; color: #666666; font-family: "Trebuchet MS", Trebuchet, Verdana, sans-serif; font-stretch: normal; line-height: normal; margin: 0.75em 0px 0px; position: relative;">
<b style="font-family: "times new roman"; line-height: 18.48px; text-align: justify;"><b style="font-family: "trebuchet ms", trebuchet, verdana, sans-serif; line-height: 27.72px;"><span style="color: blue;"><span style="font-size: large;">Related Posts</span></span></b></b></h3>
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