Neither me or a classmate I asked can figure out the answer to a couple of our questions and we can't ask the teacher easily because it is an online course.
Help or answers are appreciated.
Jess, who weighs 50 kg, is riding a 10-kg skateboard. The skateboard and Jess are moving east at 5 m/sec. Jess falls off the skateboard and hits the ground at 4 m/sec east, relative to the ground. Calculate the velocity of the skateboard.
The end of the lever of a tire jack travels 7 m for every centimeter that it lifts the car. If the car has a mass of 1000 kg and a force of 50 N is needed to lift the car, what is the efficiency of the jack?
Thank you
2006-10-26 06:46:46 · 3 answers · asked by Truthitude 1 in Science & Mathematics ➔ Physics
Thank you Potter Boy, those are explained well.
2006-10-26 07:07:12 · update #1
Question one deals with conservation of momentum.
Momentum before the event equals momentum after the event, so:
(m1+m2)v=m1v1+m2v2
(50kg+10kg)5m/s=(50kg)(4m/s)+(10kg)(v2)
300kgm/s=200kgm/s+(10kg)(v2)
100kgm/s=(10kg)(v2)
v2=10m/s
2006-10-26 07:05:38 · answer #1 · answered by disgracedfish 3 · 0⤊ 0⤋
1) conservation of momentum
(m1+m2)u = m1v1 + m2v2
(50+10)*5 = 50*4 + 10*v2
or 6*5 = 5*4 + v2
or v2 = 5*2 = 10m/s
2) for 50N force in 7m length, work done = Fx = 7*50 = 350 J
to lift against gravity, work done = mgh = 1000 * 9.8 * .01 = 10* 9.8 = 98 J
So efficiency = 98 / 350 * 100% = 28%
2006-10-26 13:55:32 · answer #2 · answered by The Potter Boy 3 · 1⤊ 0⤋
a lot of information is missing in the questions.
2006-10-26 13:51:50 · answer #3 · answered by Krishna D 2 · 0⤊ 1⤋