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Assertion: The velocity of the satellite increases as its height above earth’s surface increases and is minimum near the surface of the earth. Reason: The velocity of the satellite is directly proportional to square root of its height above earth’s

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

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Assertion: A satellite moves around the earth in a circular orbit under the action of gravity. A person in the satellite experience zero gravity field in the satellite. Reason: The contact force by the surface on the person is not

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

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Assertion: The total energy of the satellite is always negative irrespective of the nature of its orbit, i.e. elliptical or circular and it cannot be positive or zero. Reason: If the total energy is negative the satellite would leave its

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Assertion: The geo-stationary satellite goes around the earth in west-east direction. Reason Geo-stationary satellites orbits around the earth in the equatorial plane with `T=24 h` same as that of the rotation of the earth around its

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Assertion: In the satellite, everything inside it is in a state of free fall. Reason: Every part and parcel of the satellite has zero

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Assertion: An object is weightless when it is in free fall and this phenomenon is called weightlessness. Reason: In free fall, there is upward force acting on the

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 big spheres attract the nearby small ones by a force which

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 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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