I think the answer would be tensile, I’m sorry if it’s wrong
Take 10m/s^2 for the gravitational acceleration, as we know this is a free fall, we can use the equation: d=1/2*g*t^2
Substitute g=10m/s^2, t=5s, d=125m
Answer:
g = 12.22 m/s²
Explanation:
The time period of this pendulum is given as follows:
but the formula for the time period of a simple pendulum is as follows:
where,
L = length of pednulum = 48 cm = 0.48 m
g = magnitude of th gravitational acceleration on this planet = ?
Therefore,
<u>g = 12.22 m/s²</u>
Answer:
a) For the passenger the ball is seen to go up in a straight line path, and fall back to the hands of the passenger in a straight line path.
b) For a stationary observer on the ground, the ball is seen to take a parabolic path from when it is thrown up to when it fall down back on the palms of the passenger.
Explanation:
If a ball is thrown up in a vehicle moving with a constant velocity, the ball will be seen as the passenger, who is on the same frame of reference as the ball, to go up and down in a vertical straight line path. For an observer on the ground, this is different, as the ball is seen to to have both a relative vertical and horizontal component of motion, making the ball take a parabolic path from the time it was thrown, to when it falls back to the hands of the passenger.
