<h2>Since the ball rises for 2.5 s, the time to fall is 2.5 s. The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s² downward.</h2>

6 0

3 years ago

Which would deliver a greater change in momentum to an opponent’s body - a dodgeball that traveled at 10m/s and rebounded with a

Mazyrski [523]

2nd sentence seems bout right ?? i think

7 0

3 years ago

Suppose that you are standing on a train accelerating at 0.20g (where g is the acceleration due to gravity). What minimum coeffi

prisoha [69]

Answer:

0.2

Explanation:

The given parameters are;

The acceleration of the train, a = 0.2·g

The mass of the person standing on the train = m

Let μ represent the coefficient of static friction, we have;

The force acting on the person, F = m × a = m × 0.2·g

The force of friction acting between the feet and the floor, $F_f$ = m·g·μ

For the person not to slide we have;

The force acting on the person = The force of friction acting between the feet and the floor

F = $F_f$

∴ m × 0.2·g = m·g·μ

From which we get;

0.2 = μ

The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.

7 0

3 years ago