Acceleration in 2D
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A mass falls from a height ‘h’ and its time of fall ‘t’ is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'. Then we can say
A particle hanging from a spring stretches it by 1 cm at earth's surface. How much will the same particle stretch the spring at a place 800 km above the earth's surface? Radius of the earth = 6400 km.
1.2cm
0.79 cm.
0.29 cm.
0.99 cm.
- 6 ms−2
- 8 ms−2
- 10 ms−2
- 14 ms−2
A particle moves along a straight line according to the law , where is the distance of the particle from a fixed point at the end of . The acceleration of the particle at the end of is
None of these