A(n) 0.49 kg softball is pitched at a speed of 13 m/s. the batter hits it back directly at the pitcher at a speed of 23 m/s. the bat acts on the ball for 0.014 s. what is the magnitude of the impulse imparted by the bat to the ball?

Explanation:

Impulse is given by change of momentum.

Mass of softball = 0.49 kg

Initial velocity of softball = 13 m/s

Initial momentum = 0.49 x 13 = 6.37 kgm/s

Final velocity of softball = 23 m/s

Final momentum = 0.49 x 23 = 11.27 kgm/s

Change of momentum = Final momentum - Initial momentum

Change of momentum = 11.27 - 6.37

Change of momentum = 4.9 kgm/s

Impulse = Change of momentum

Impulse  = 4.9 kgm/s

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6