Answer to Question #169013 in Physics for Jennifer

Question #169013

1)What net force is required to increase the velocity of a 115 kg scooter from 4.50 m/s to 18.0 m/s over a time of 12.0 s?

2) A 131 kg NFL defensive end runs to the right at 7.80 m/s towards a 97.0 kg running back that is running at 8.00 m/s to the left. The players have a head-on, perfectly inelastic collision. What is the velocity of players after the defensive end makes the tackle?

3)An 85.0 kg bicycle and its rider riding at 9.00 m/s to the right crashes into a 56.0 kg skater that is moving at 3.00 m/s to the left. After the crash, the bicycle and rider continue moving to the right at 1.20 m/s to the right. What is the final velocity of the skater?

4)Two boxes on a frictionless surface have a head-on, *perfectly elastic* collision.


Box 1 has a mass of 3.00 kg and an initial velocity of 5.00 m/s to the right.

Box 2 has a mass of 8.00 kg and an initial velocity of 2.00 m/s to the left.


What are the velocities of both boxes after they collide?

Write your answer in a complete sentence.



1
Expert's answer
2021-03-04T17:24:12-0500

1)


F=115174.512=120 NF=115\frac{17-4.5}{12}=120\ N

2)


(131)(7.8)(97)(8)=(131+97)vv=1.08ms(131)(7.8)-(97)(8)=(131+97)v\\v=1.08\frac{m}{s}

3)


(85)(9)(56)(3)=(85)(1.2)+(56)vv=8.8ms(85)(9)-(56)(3)=(85)(1.2)+(56)v\\v=8.8\frac{m}{s}

4)


v1f=383+852283+8=5.2msv2f=233+852833+8=1.8msv_{1f}=\frac{3-8}{3+8}5-2\frac{2\cdot8}{3+8}=-5.2\frac{m}{s}\\ v_{2f}=\frac{2\cdot3}{3+8}5-2\frac{8-3}{3+8}=1.8\frac{m}{s}


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