I found the answer for you if u need any help ask anytime!
From kinematic relation d = vi t + 1/2 at^2 , a = 2d/t^2 since it starts from rest . Acceleration does not depend on the mass. a = 2* 2.5 / (1.47)^2 = 2.3139 m/s^2.
Acceleration down the slide is due to the parallel component of gravity force. Friction opposes this force and acts up the inclined plane. mg sin theta - f = ma. Friction = f = (2.45 * 9.8) sin 30.5 - (2.45)(2.3139) = 6.5169 N.
Normal force = mg cos 30.5 = (2.45*9.8)* cos 30.5 = 20.6877 N. Friction = uk N. coefficient of friction = = friction / N = 6.5169 / 20.6877 = 0.32.
When it reaches the bottom of the slide, vf = vi + at = 0 + ( 2.3139)(1.47) = 3.4014 m/s = velocity at the bottom
Answer:
Work done, W = 6 J
Explanation:
It is given that,
Force of gravity acting on the book, weight of the book is 15 N
We need to find the work done in lifting the book straight up for a distance of 0.4 meters.
The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :
So, the magnitude of work done in lifting the book is 6 joules.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer: is C
Expiation: you can find the answer on quizlet and it’s diverging mirror so the answer is C