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

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

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