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Cavendish’s ExperimentThe figure shows the schematic drawing of Cavendish’s experiment to determine the value of the gravitational constant. The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.Two large lead spheres are brought close to the small ones but on opposite sides as shown. The value of G from this experiment came to be `6.67 xx 10^(-11) N-m^2'/'kg^2`.The net force on the bar

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Cavendish’s ExperimentThe figure shows the schematic drawing of Cavendish’s experiment to determine the value of the gravitational constant. The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.Two large lead spheres are brought close to the small ones but on opposite sides as shown. The value of G from this experiment came to be `6.67 xx 10^(-11) N-m^2'/'kg^2`.The net torque on the bar

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Cavendish’s ExperimentThe figure shows the schematic drawing of Cavendish’s experiment to determine the value of the gravitational constant. The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.Two large lead spheres are brought close to the small ones but on opposite sides as shown. The value of G from this experiment came to be `6.67 xx 10^(-11) N-m^2'/'kg^2`.The torque produces twist in the suspended wire. The twisting stops

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Cavendish’s ExperimentThe figure shows the schematic drawing of Cavendish’s experiment to determine the value of the gravitational constant. The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.Two large lead spheres are brought close to the small ones but on opposite sides as shown. The value of G from this experiment came to be `6.67 xx 10^(-11) N-m^2'/'kg^2`.After Cavendish’s experiment, there have been given suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years)

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Acceleration due to gravityThe acceleration for any object moving under the sole influence of gravity is known as acceleration due to gravity. So, for an object of mass `m,` the acceleration experienced by it is usually denoted by the symbol g which is related to F by Newton’s second law by relation `F= mg`. Thus,`g=F/m=(GM_e)/(r_e^2)`Acceleration `g` is readily measurable as `R_e` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g and R_e` enables one to estimate `M_e` from the above equation. This is the reason why there is a popular statement

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Acceleration due to gravityThe acceleration for any object moving under the sole influence of gravity is known as acceleration due to gravity. So, for an object of mass `m,` the acceleration experienced by it is usually denoted by the symbol g which is related to F by Newton’s second law by relation `F= mg`. Thus,`g=F/m=(GM_e)/(r_e^2)`Acceleration `g` is readily measurable as `R_e` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g and R_e` enables one to estimate `M_e` from the above equation. This is the reason why there is a popular statement

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Acceleration due to gravityThe acceleration for any object moving under the sole influence of gravity is known as acceleration due to gravity. So, for an object of mass `m,` the acceleration experienced by it is usually denoted by the symbol g which is related to F by Newton’s second law by relation `F= mg`. Thus,`g=F/m=(GM_e)/(r_e^2)`Acceleration `g` is readily measurable as `R_e` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g and R_e` enables one to estimate `M_e` from the above equation. This is the reason why there is a popular statement

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Acceleration due to gravityThe acceleration for any object moving under the sole influence of gravity is known as acceleration due to gravity. So, for an object of mass `m,` the acceleration experienced by it is usually denoted by the symbol g which is related to F by Newton’s second law by relation `F= mg`. Thus,`g=F/m=(GM_e)/(r_e^2)`Acceleration `g` is readily measurable as `R_e` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g and R_e` enables one to estimate `M_e` from the above equation. This is the reason why there is a popular statement

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Acceleration due to gravityThe acceleration for any object moving under the sole influence of gravity is known as acceleration due to gravity. So, for an object of mass `m,` the acceleration experienced by it is usually denoted by the symbol g which is related to F by Newton’s second law by relation `F= mg`. Thus,`g=F/m=(GM_e)/(r_e^2)`Acceleration `g` is readily measurable as `R_e` is a known quantity. The measurement of `G` by Cavendish’s experiment (or otherwise), combined with knowledge of `g and R_e` enables one to estimate `M_e` from the above equation. This is the reason why there is a popular statement

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Earth’s SatelliteEarth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Earth’s SatelliteEarth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Earth’s SatelliteEarth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Earth’s SatelliteEarth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Earth’s SatelliteEarth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Which of the following examples represent periodic

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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The displacement of a particle is represented by the equation `y = 3 cos (pi/ 4 - ωt)` . The motion of the particle

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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The displacement of a particle is represented by the equation `y = sin^3 ωt` . The motion

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The relation between acceleration and displacement of four particles are given

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Motion of an oscillating liquid column in a U-tube

asked Mar 19, 2022 in 11th Physics by varun (6.7k points)

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