q9 An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graphof the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After thecollision, the 2 kg object has a speed of 2 m, and the 1 kg object has a speed of 8 m. A student mustdetermine the initial speed vo of the 2 kg object. Which of the following options shows the correctsubstitutions into the equation Ap = FAt so that the initial speed of the unknown object can bedetermined?(2 kg)(2 m – vo) = (8000 N)(0.00125 s), because the maximum force exerted onthe 2 kg object must be considered.(2 kg)(2 - vo)(8000 N)(0.00125 s), because the average force exerted onmSthe 2 kg object must be considered.(2 kg – 1 kg)(2 -vo)(8000 N)(0.00125 s), because the difference in massand the maximum force exerted on the 2 kg object must be considered.(2 kg – 1 kg)(2 m/s – vo) = ;(8000 N)(0.00125 s), because the difference inmass and the average force exerted on the 2 kg object must be considered. +y>+X2 kg1 kg10,00080002 6000400020000 -0.00100.00150.00200.0025t (s)

Given- Mass of object 1 (m1) = 2 Kg, Mass of object 2 (m2) = 1 Kg, As object 2 is in rest initially,…

An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graph of the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After the collision, the 2 kg object has a speed of 2 , and the 1 kg object has a speed of 8 . A student must determine the initial speed vo of the 2 kg object. Which of the following options shows the correct substitutions into the equation Ap = FAt so that the initial speed of the unknown object can be determined? (2 kg)(2 - vo) = (8000 N)(0.00125 s), because the maximum force exerted on the 2 kg object must be considered. (2 kg)(2 – vo) = ÷(8000 N)(0.00125 s), because the average force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 -vo) = (8000 N)(0.00125 s), because the difference in mass and the maximum force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 m/s – vo) = (8000 N)(0.00125 s), because the difference in mass and the average force exerted on the 2 kg object must be considered.

An object of mass 2 kg collides with an object of mass 1 kg that is at rest, as shown in the figure. A graph of the force as a function of time that the 1 kg object exerts on the 2 kg object is shown. After the collision, the 2 kg object has a speed of 2 , and the 1 kg object has a speed of 8 . A student must determine the initial speed vo of the 2 kg object. Which of the following options shows the correct substitutions into the equation Ap = FAt so that the initial speed of the unknown object can be determined? (2 kg)(2 - vo) = (8000 N)(0.00125 s), because the maximum force exerted on the 2 kg object must be considered. (2 kg)(2 – vo) = ÷(8000 N)(0.00125 s), because the average force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 -vo) = (8000 N)(0.00125 s), because the difference in mass and the maximum force exerted on the 2 kg object must be considered. (2 kg – 1 kg)(2 m/s – vo) = (8000 N)(0.00125 s), because the difference in mass and the average force exerted on the 2 kg object must be considered.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

Q8: Study the diagram carefully and ignore air resistance. v, = 22.7 m/s 01 = 66 ° 02 = 45° %3D y m Vy Vox Vox 4.5 V0x voy Vox x m V0x voy Find: (a) vy in position 4m, (b) vz in position 14m, (c) vy in position 14m, (d) vy just before strike the ground.
46. A 2.55 kg bucket of sand is attached to a 1.31 m long rope and is swung in a vertical circle. At the bottom of the circle the tension in the rope is 30.2 N. What is the speed of the bucket of sand at the bottom of the circle? A. 1.85 m/sB. 2.61 m/sC. 3.41 m/sD. 6.01 m/s Please choose one of these answers!!!!
Q3
14.
Answer Q8. Hence Justify your answer
3. How do you graphically represent the work done on an object moved from d, to d,? Show this on the graph
Q3 please again
no. 2 e
I watched a lot of Looney Toons when I was a kid, and liked that the Roadrunners legs would spin for a second or two before their claws caught traction. a. With a coefficient of kinetic friction 0.2 what acceleration will the roadrunner experience while their feet are sliding accross the ground? b. How far forward will the roadrunner drift in a half second, starting from rest under this acceleration? c. After catching traction the Roadrunner accelerates away, travelling 10 m to get offscreen in one fifth of a second. What was their coefficient of static friction?
Q A & QB PLEASE
9. A motorboat travels across a 132 foot wide river at 7 mi/h directed 18 degrees into a 3 mi/h current. What is the velocity of the boat relative to the bank? 18%
In the figure below, find T1 and T2