355638 The earth circles the sun once a year. How much work would have to be done on the earth to bring it to rest relative to the sun, (ignore the rotation of earth about-its own axis) Given that mass of the earth is \(6 \times {10^{24}}\;kg\) and distance between the sun and earth is \(1.5 \times {10^{11}}\;m\)

355639 A particle moves on a rough horizontal ground with some initial velocity say \({v_0}\). If \(\left(\dfrac{3}{4}\right)^{\text {th }}\) of its kinetic energy is lost due to friction in time \({t_0}\) then coefficient of friction between the particle and the ground is

355640 A block of mass \(10\;kg\) is moving in \(x\)-direction with a constant speed of \(10\;m\;{s^{ - 1}}\). It is subjected to a retarding force \({F_r} = - 0.1x\;J\;{m^{ - 1}}\) during its travel from \(x = 20\;m\) to \(x = 30\;m\). Its final kinetic energy will be

355641 A bullet of mass \(100\;g\) moving horizontally with a velocity \(210\;m{s^{ - 1}}\) gets embedded in a block of mass \(2\;kg\) kept on a rough horizontal surface. If the coefficient of kinetic friction between the block and surface is \(0.5.\) Calculate the distance block - bullet system will move before coming to rest.
(Take acceleration due to gravity \( = 10\;m{s^{ - 2}}\)).

355642 A variable force, given by the 2-dimensional vector \(\vec{F}=\left(3 x^{2} \hat{i}+4 \hat{j}\right)\), acts on a particle. The force is in newton and \(\mathrm{x}\) is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates \((2,3)\) to \((3,0)\) ( the coordinates are in metres)

355638 The earth circles the sun once a year. How much work would have to be done on the earth to bring it to rest relative to the sun, (ignore the rotation of earth about-its own axis) Given that mass of the earth is \(6 \times {10^{24}}\;kg\) and distance between the sun and earth is \(1.5 \times {10^{11}}\;m\)

355639 A particle moves on a rough horizontal ground with some initial velocity say \({v_0}\). If \(\left(\dfrac{3}{4}\right)^{\text {th }}\) of its kinetic energy is lost due to friction in time \({t_0}\) then coefficient of friction between the particle and the ground is

355640 A block of mass \(10\;kg\) is moving in \(x\)-direction with a constant speed of \(10\;m\;{s^{ - 1}}\). It is subjected to a retarding force \({F_r} = - 0.1x\;J\;{m^{ - 1}}\) during its travel from \(x = 20\;m\) to \(x = 30\;m\). Its final kinetic energy will be

355641 A bullet of mass \(100\;g\) moving horizontally with a velocity \(210\;m{s^{ - 1}}\) gets embedded in a block of mass \(2\;kg\) kept on a rough horizontal surface. If the coefficient of kinetic friction between the block and surface is \(0.5.\) Calculate the distance block - bullet system will move before coming to rest.
(Take acceleration due to gravity \( = 10\;m{s^{ - 2}}\)).

355642 A variable force, given by the 2-dimensional vector \(\vec{F}=\left(3 x^{2} \hat{i}+4 \hat{j}\right)\), acts on a particle. The force is in newton and \(\mathrm{x}\) is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates \((2,3)\) to \((3,0)\) ( the coordinates are in metres)

355638 The earth circles the sun once a year. How much work would have to be done on the earth to bring it to rest relative to the sun, (ignore the rotation of earth about-its own axis) Given that mass of the earth is \(6 \times {10^{24}}\;kg\) and distance between the sun and earth is \(1.5 \times {10^{11}}\;m\)

355639 A particle moves on a rough horizontal ground with some initial velocity say \({v_0}\). If \(\left(\dfrac{3}{4}\right)^{\text {th }}\) of its kinetic energy is lost due to friction in time \({t_0}\) then coefficient of friction between the particle and the ground is

355640 A block of mass \(10\;kg\) is moving in \(x\)-direction with a constant speed of \(10\;m\;{s^{ - 1}}\). It is subjected to a retarding force \({F_r} = - 0.1x\;J\;{m^{ - 1}}\) during its travel from \(x = 20\;m\) to \(x = 30\;m\). Its final kinetic energy will be

355641 A bullet of mass \(100\;g\) moving horizontally with a velocity \(210\;m{s^{ - 1}}\) gets embedded in a block of mass \(2\;kg\) kept on a rough horizontal surface. If the coefficient of kinetic friction between the block and surface is \(0.5.\) Calculate the distance block - bullet system will move before coming to rest.
(Take acceleration due to gravity \( = 10\;m{s^{ - 2}}\)).

355642 A variable force, given by the 2-dimensional vector \(\vec{F}=\left(3 x^{2} \hat{i}+4 \hat{j}\right)\), acts on a particle. The force is in newton and \(\mathrm{x}\) is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates \((2,3)\) to \((3,0)\) ( the coordinates are in metres)

355638 The earth circles the sun once a year. How much work would have to be done on the earth to bring it to rest relative to the sun, (ignore the rotation of earth about-its own axis) Given that mass of the earth is \(6 \times {10^{24}}\;kg\) and distance between the sun and earth is \(1.5 \times {10^{11}}\;m\)

355639 A particle moves on a rough horizontal ground with some initial velocity say \({v_0}\). If \(\left(\dfrac{3}{4}\right)^{\text {th }}\) of its kinetic energy is lost due to friction in time \({t_0}\) then coefficient of friction between the particle and the ground is

355640 A block of mass \(10\;kg\) is moving in \(x\)-direction with a constant speed of \(10\;m\;{s^{ - 1}}\). It is subjected to a retarding force \({F_r} = - 0.1x\;J\;{m^{ - 1}}\) during its travel from \(x = 20\;m\) to \(x = 30\;m\). Its final kinetic energy will be

355641 A bullet of mass \(100\;g\) moving horizontally with a velocity \(210\;m{s^{ - 1}}\) gets embedded in a block of mass \(2\;kg\) kept on a rough horizontal surface. If the coefficient of kinetic friction between the block and surface is \(0.5.\) Calculate the distance block - bullet system will move before coming to rest.
(Take acceleration due to gravity \( = 10\;m{s^{ - 2}}\)).

355642 A variable force, given by the 2-dimensional vector \(\vec{F}=\left(3 x^{2} \hat{i}+4 \hat{j}\right)\), acts on a particle. The force is in newton and \(\mathrm{x}\) is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates \((2,3)\) to \((3,0)\) ( the coordinates are in metres)

355638 The earth circles the sun once a year. How much work would have to be done on the earth to bring it to rest relative to the sun, (ignore the rotation of earth about-its own axis) Given that mass of the earth is \(6 \times {10^{24}}\;kg\) and distance between the sun and earth is \(1.5 \times {10^{11}}\;m\)

355639 A particle moves on a rough horizontal ground with some initial velocity say \({v_0}\). If \(\left(\dfrac{3}{4}\right)^{\text {th }}\) of its kinetic energy is lost due to friction in time \({t_0}\) then coefficient of friction between the particle and the ground is

355640 A block of mass \(10\;kg\) is moving in \(x\)-direction with a constant speed of \(10\;m\;{s^{ - 1}}\). It is subjected to a retarding force \({F_r} = - 0.1x\;J\;{m^{ - 1}}\) during its travel from \(x = 20\;m\) to \(x = 30\;m\). Its final kinetic energy will be

355641 A bullet of mass \(100\;g\) moving horizontally with a velocity \(210\;m{s^{ - 1}}\) gets embedded in a block of mass \(2\;kg\) kept on a rough horizontal surface. If the coefficient of kinetic friction between the block and surface is \(0.5.\) Calculate the distance block - bullet system will move before coming to rest.
(Take acceleration due to gravity \( = 10\;m{s^{ - 2}}\)).

355642 A variable force, given by the 2-dimensional vector \(\vec{F}=\left(3 x^{2} \hat{i}+4 \hat{j}\right)\), acts on a particle. The force is in newton and \(\mathrm{x}\) is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates \((2,3)\) to \((3,0)\) ( the coordinates are in metres)