Physics

A 0.060kg ball is propelled by a 1.10kg racquet from rest to a speed of 40.0m/s. If the two objects are in contact for 250ms, calculate the average force exerted by the racquet on the ball.

Feedback: F*t = m*v F = m*v/t = 0.06*40/0.25 = 2.4/0.25 = 9.60N

A 60.0kg swimmer stands at the edge of a 200kg boat that is motionless on the water of a lake. If the swimmer dives from the boat in a horizontal direction due east at 8.00m/s, calculate the velocity of the boat if friction between the boat and the water is neglected.

Feedback: po = pfms*vso+mb*vbo = ms*vsf+mb*vbf60*0+200*0 = 60*8+200*vbfvbf=-480/200=-2.40m/s (opposite to direction of swimmer)

A 500kg car traveling due east at 54.0km/h collides head on with a 3000kg van moving at 18.0km/h. If the two vehicles stick to each other as a result of the collision, calculate the velocity of the combined mass. Feedback: po = pf mc*vco+mv*vvo = mc*vcf+mv*vvf 500*15+3000*(-5 )= (500+3000)*vf vf = -7,500/3,500 = -2.14m/s at 0 degrees (due east) or 2.14m/s at 180 degrees (due west, same direction as van)

An 18.0g arrow is fired into a 3.00kg pumpkin suspended on a string. The arrow imbeds itself in the pumpkin and makes it move at 1.50m/s due east. Calculate the velocity of the arrow just before it enters the pumpkin. Feedback: po = pf ma*vao+mp*vpo = (ma+mp)*vf 0.018*vao+3*0 = 3.018*1.5 vao = 4.527/0.018 = 252m/s

#9 Assuming that the earth is a perfect sphere with a radius of 6.38Mm, calculate the speed and the centripetal acceleration of a person standing on the equator. Calculate the same for a person at latitude 45 degrees.Correct Answer(s): om=1rev/day*1day/24h*1h/3600s*2 pi rad/rev=7.27*10^(-5)rad/s vt=r*om=6.38*10^6*7.27*10^(-5)=464m/s ac=vt^2/r=r*om^2=6.38*10^6*7.27^2*10^(-10)=0.0337m/s/s

at 45 degrees r=6.38*10^6*cos45=4.511*10^6 vt=328m/s ac=4.511*10^6*7.27^2*10(-10)=0.0239m/s^2

The coefficient of friction between the road and the tires of a 900kg car is 0.888. Calculate the maximum speed for the car...