Now assume that the pitcher in part d throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. the batter then hits the ball so it goes directly back to the pitcher along the same straight line. what is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4n⋅s to the baseball?

25.9 m/s

Explanation:

mass of ball, m = 0.145 kg

initial velocity, u = + 32 m/s

It bounce back with the velocity but in opposite direction so final velocity,

v = - v

Impulse, I = - 8.4 Ns

Impulse is defined as the change in momentum

I = m v - mu = m (v - u)

- 8.4 = 0.145 x (- v - 3 2)

- 57.9 = - v - 32

v = 57.9 - 32 = 25.9 m/s

Thus, the final speed of the ball is 25.9 m/s

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation: