A bat strikes a 50 g baseball so that its velocity changes by + 30 m/s in 0.10 s

Question

A bat strikes a 50 g baseball so that its velocity changes by + 30 m/s in 0.10 s. With what average force was the ball struck? +15 N +300 N +150 N -15 N A 2 kg pistol fires a 1 g bullet with a muzzle speed of 1000 m/s. The bullet strikes a 10 kg wooden block that rests on a horizontal frictionless surface. The block, with the embedded bullet, then sides along the surface. What is the kinetic energy of the bullet as it travels toward the block? 1.0 J 1000 J 10.000 J 500 J The explosive charge in the pistol acts for 0.001 s. What is the average force exerted on the bullet while it is being fired 0.001 N 100 N 1000 N 1.0 N 500 N What is the speed of the "bullet + block" system immediately after the bullet is embedded in the block? 0.1 m/s 1000 m/s zero 10 m/s A grindstone of radius 4 m is initially spinning with a constant angular speed of 8.0 rad/s. The angular speed is the increased to 10 rad/s the next 4 seconds. Assume that the angular acceleration is constant. What is the average angular speed of the grindstone? 0.5 rad/s 4.5 rad/s 18 rad/s 2.0 rad/s 9.0 rad/s What is the magnitude of the angular acceleration of the grindstone? 0.5 rad/s^2 4.5 rad/s^2 Through how many revolutions does the grindstone turn during the 4 second interval? 0.64 4.00 36.0 3.82 What is the angular speed, in rad/s, of the second hand of a watch? 1.7 Times 10^-3 0.02 60 0.10 6.28

Explanation / Answer

(25).mass m = 50 g = 0.05 kg

Change in velocity v = 30 m/s

time t = 0.1 s

Average force F = ma

                       = mv/t

                       = 0.05 x30 / 0.1

                       = 15 N

(26) mass of pistol M = 2kg

mass of bullet m = 1 g = 0.001 kg

muzzle speed u = 1000 m/s

mass of wooden block M ' = 10 kg

Kinetic energy of the bullet K.E = (1/2) mu 2

                                              = 0.5 x0.001 x1000 2

                                              = 500 J

(27) Required force = mu / t

                            = (0.001)(1000)/0.001

                            = 1000 N

(28) From law of conservation of momentum ,

mu + M' U = (m+M ' )v

            mu = (m+ M ' ) v

From this v = mu /(m+M ')

                  = (0.001x1000) /(0.001+10)

                  = 99.99 x10 -3 m/s

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