270472
A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is
1 \(\mathrm{V}=\sqrt{2 g R}\)
2 \(\mathrm{V}=3 \mathrm{gR}\)
3 \(\mathrm{V}=\sqrt{g R}\)
4 \(\mathrm{V}=2 \mathrm{gR}\)
Explanation:
Gravitation
270473
When projectile attains escape velocity, then on the surface of planet, its
1 \(K E\lt P E\)
2 \(P E\lt K E\)
3 \(K E=P E\)
4 \(K E=2 P E\)
Explanation:
Gravitation
270474
A satellite is moving with constant speed ' \(V\) ' in a circular orbit around earth. The kinetic energy of the satellite is
1 \(\frac{1}{2} m V^{2}\)
2 \(m V^{2}\)
3 \(\frac{3}{2} m V^{2}\)
4 \(2 m V^{2}\)
Explanation:
Gravitation
270528
A satellite moves around the earth in a circular orbit with speed ' \(v\) '. If ' \(m\) ' is mass of the satellite then its total energy is
1 \(\frac{1}{2} \mathrm{mv}^{2}\)
2 \(\mathrm{mv}^{2}\)
3 \(-\frac{1}{2} m v^{2}\)
4 \(\frac{3}{2} m v^{2}\)
Explanation:
TE \(=-\mathrm{KE}=-\frac{1}{2} m v^{2}\)
Gravitation
270529
The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.
1 \(2 \mathrm{E}\)
2 \(4 \mathrm{E}\)
3 \(2 \sqrt{2} E\)
4 \(\sqrt{2} \mathrm{E}\)
Explanation:
\(\frac{K_{e}}{K_{0}}=\frac{2 g R}{g R} \Rightarrow K_{e}=2 K_{0}\)
270472
A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is
1 \(\mathrm{V}=\sqrt{2 g R}\)
2 \(\mathrm{V}=3 \mathrm{gR}\)
3 \(\mathrm{V}=\sqrt{g R}\)
4 \(\mathrm{V}=2 \mathrm{gR}\)
Explanation:
Gravitation
270473
When projectile attains escape velocity, then on the surface of planet, its
1 \(K E\lt P E\)
2 \(P E\lt K E\)
3 \(K E=P E\)
4 \(K E=2 P E\)
Explanation:
Gravitation
270474
A satellite is moving with constant speed ' \(V\) ' in a circular orbit around earth. The kinetic energy of the satellite is
1 \(\frac{1}{2} m V^{2}\)
2 \(m V^{2}\)
3 \(\frac{3}{2} m V^{2}\)
4 \(2 m V^{2}\)
Explanation:
Gravitation
270528
A satellite moves around the earth in a circular orbit with speed ' \(v\) '. If ' \(m\) ' is mass of the satellite then its total energy is
1 \(\frac{1}{2} \mathrm{mv}^{2}\)
2 \(\mathrm{mv}^{2}\)
3 \(-\frac{1}{2} m v^{2}\)
4 \(\frac{3}{2} m v^{2}\)
Explanation:
TE \(=-\mathrm{KE}=-\frac{1}{2} m v^{2}\)
Gravitation
270529
The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.
1 \(2 \mathrm{E}\)
2 \(4 \mathrm{E}\)
3 \(2 \sqrt{2} E\)
4 \(\sqrt{2} \mathrm{E}\)
Explanation:
\(\frac{K_{e}}{K_{0}}=\frac{2 g R}{g R} \Rightarrow K_{e}=2 K_{0}\)
270472
A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is
1 \(\mathrm{V}=\sqrt{2 g R}\)
2 \(\mathrm{V}=3 \mathrm{gR}\)
3 \(\mathrm{V}=\sqrt{g R}\)
4 \(\mathrm{V}=2 \mathrm{gR}\)
Explanation:
Gravitation
270473
When projectile attains escape velocity, then on the surface of planet, its
1 \(K E\lt P E\)
2 \(P E\lt K E\)
3 \(K E=P E\)
4 \(K E=2 P E\)
Explanation:
Gravitation
270474
A satellite is moving with constant speed ' \(V\) ' in a circular orbit around earth. The kinetic energy of the satellite is
1 \(\frac{1}{2} m V^{2}\)
2 \(m V^{2}\)
3 \(\frac{3}{2} m V^{2}\)
4 \(2 m V^{2}\)
Explanation:
Gravitation
270528
A satellite moves around the earth in a circular orbit with speed ' \(v\) '. If ' \(m\) ' is mass of the satellite then its total energy is
1 \(\frac{1}{2} \mathrm{mv}^{2}\)
2 \(\mathrm{mv}^{2}\)
3 \(-\frac{1}{2} m v^{2}\)
4 \(\frac{3}{2} m v^{2}\)
Explanation:
TE \(=-\mathrm{KE}=-\frac{1}{2} m v^{2}\)
Gravitation
270529
The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.
1 \(2 \mathrm{E}\)
2 \(4 \mathrm{E}\)
3 \(2 \sqrt{2} E\)
4 \(\sqrt{2} \mathrm{E}\)
Explanation:
\(\frac{K_{e}}{K_{0}}=\frac{2 g R}{g R} \Rightarrow K_{e}=2 K_{0}\)
270472
A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is
1 \(\mathrm{V}=\sqrt{2 g R}\)
2 \(\mathrm{V}=3 \mathrm{gR}\)
3 \(\mathrm{V}=\sqrt{g R}\)
4 \(\mathrm{V}=2 \mathrm{gR}\)
Explanation:
Gravitation
270473
When projectile attains escape velocity, then on the surface of planet, its
1 \(K E\lt P E\)
2 \(P E\lt K E\)
3 \(K E=P E\)
4 \(K E=2 P E\)
Explanation:
Gravitation
270474
A satellite is moving with constant speed ' \(V\) ' in a circular orbit around earth. The kinetic energy of the satellite is
1 \(\frac{1}{2} m V^{2}\)
2 \(m V^{2}\)
3 \(\frac{3}{2} m V^{2}\)
4 \(2 m V^{2}\)
Explanation:
Gravitation
270528
A satellite moves around the earth in a circular orbit with speed ' \(v\) '. If ' \(m\) ' is mass of the satellite then its total energy is
1 \(\frac{1}{2} \mathrm{mv}^{2}\)
2 \(\mathrm{mv}^{2}\)
3 \(-\frac{1}{2} m v^{2}\)
4 \(\frac{3}{2} m v^{2}\)
Explanation:
TE \(=-\mathrm{KE}=-\frac{1}{2} m v^{2}\)
Gravitation
270529
The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.
1 \(2 \mathrm{E}\)
2 \(4 \mathrm{E}\)
3 \(2 \sqrt{2} E\)
4 \(\sqrt{2} \mathrm{E}\)
Explanation:
\(\frac{K_{e}}{K_{0}}=\frac{2 g R}{g R} \Rightarrow K_{e}=2 K_{0}\)
270472
A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is
1 \(\mathrm{V}=\sqrt{2 g R}\)
2 \(\mathrm{V}=3 \mathrm{gR}\)
3 \(\mathrm{V}=\sqrt{g R}\)
4 \(\mathrm{V}=2 \mathrm{gR}\)
Explanation:
Gravitation
270473
When projectile attains escape velocity, then on the surface of planet, its
1 \(K E\lt P E\)
2 \(P E\lt K E\)
3 \(K E=P E\)
4 \(K E=2 P E\)
Explanation:
Gravitation
270474
A satellite is moving with constant speed ' \(V\) ' in a circular orbit around earth. The kinetic energy of the satellite is
1 \(\frac{1}{2} m V^{2}\)
2 \(m V^{2}\)
3 \(\frac{3}{2} m V^{2}\)
4 \(2 m V^{2}\)
Explanation:
Gravitation
270528
A satellite moves around the earth in a circular orbit with speed ' \(v\) '. If ' \(m\) ' is mass of the satellite then its total energy is
1 \(\frac{1}{2} \mathrm{mv}^{2}\)
2 \(\mathrm{mv}^{2}\)
3 \(-\frac{1}{2} m v^{2}\)
4 \(\frac{3}{2} m v^{2}\)
Explanation:
TE \(=-\mathrm{KE}=-\frac{1}{2} m v^{2}\)
Gravitation
270529
The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.
1 \(2 \mathrm{E}\)
2 \(4 \mathrm{E}\)
3 \(2 \sqrt{2} E\)
4 \(\sqrt{2} \mathrm{E}\)
Explanation:
\(\frac{K_{e}}{K_{0}}=\frac{2 g R}{g R} \Rightarrow K_{e}=2 K_{0}\)
