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An object of mass 0.2 kg executes SHM along the x-axis with frequency of (25/pie) Hz. At the point x = 0.04 m, the object has kinetic energy 0.5 J and potential energy 0.4 J.
30
Oct
An object of mass 0.2 kg executes SHM along the x-axis with frequency of (25/pie) Hz. At the point x = 0.04 m, the object has kinetic energy 0.5 J and potential energy 0.4 J. An object of mass 0.2 kg executes SHM along the x-axis with frequency of (25/pie) Hz. At the point x [...]
The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if it is a second’s pendulum on earth)
30
Oct
The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if it is a second’s pendulum on earth) The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if [...]
The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, find its time period.
30
Oct
The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, find its time period. find its time period. The potential energy of a simple harmonic oscillator of mass 2 kg in its mean [...]
Calculate the total electric flux through the paraboloidal surface due to a uniform electric field of magnitude E0 in the direction shown in figure.
30
Oct
Calculate the total electric flux through the paraboloidal surface due to a uniform electric field of magnitude E0 in the direction shown in figure. Calculate the total electric flux through the paraboloidal surface due to a uniform electric field of magnitude E0 in the direction shown in figure. October 30, 2020 Category: Uncategorised (JEE Advanced [...]
A body performs simple harmonic oscillation along the straight line PQRST with R as the midpoint of PT. Its kinetic energies at Q and S are each one fourth of its maximum value.
30
Oct
A body performs simple harmonic oscillation along the straight line PQRST with R as the midpoint of PT. Its kinetic energies at Q and S are each one fourth of its maximum value. A body performs simple harmonic oscillation along the straight line PQRST with R as the midpoint of PT. Its kinetic energies at [...]
A pendulum clock, which keeps correct time at sea level, loses 15s per day when taken to the top of a mountain. If the radius of the earth is 6400 km, the height of the mountain is
30
Oct
A pendulum clock, which keeps correct time at sea level, loses 15s per day when taken to the top of a mountain. If the radius of the earth is 6400 km, the height of the mountain is A pendulum clock loses 15s per day when taken to the top of a mountain. If the radius [...]
A pendulum is taken1 km inside from sea level. It
30
Oct
A pendulum is taken1 km inside from sea level. It A pendulum is taken1 km inside from sea level. It October 30, 2020 Category: Cengage NEET by C.P Singh , Chapter 13 - Simple Harmonic Motion , Part 1 ,
A particle starts oscillating simple harmonically with time period T from its equilibrium position. Find the ratio of kinetic energy and potential energy of the particle at the time
30
Oct
A particle starts oscillating simple harmonically with time period T from its equilibrium position. Find the ratio of kinetic energy and potential energy of the particle at the time A particle starts oscillating simple harmonically with time period T from its equilibrium position. Find the ratio of kinetic energy and potential energy of the particle [...]
A uniform electric field ai + bj intersects a surface of area A. what is the flux through this area if the surface lies ( a) in the yz plane , ( b) in the xz plane ( c) in the xy plane?
30
Oct
A uniform electric field ai + bj intersects a surface of area A. what is the flux through this area if the surface lies ( a) in the yz plane , ( b) in the xz plane ( c) in the xy plane? ( b) in the xz plane ( c) in the xy plane? [...]
The motion of a particle is described by x = 30 sin (pie t + pie / 6) , where x is in cm and t in sec. Find the position where potential energy of the particle is twice of kinetic energy
30
Oct
The motion of a particle is described by x = 30 sin (pie t + pie / 6) , where x is in cm and t in sec. Find the position where potential energy of the particle is twice of kinetic energy The motion of a particle is described by x = 30 sin (pie [...]