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microwave13016
30.03.2020 •
Physics
A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, how long will it take (in s) before the box moves without slipping
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Ответ:
t = 1.16 s.
Explanation:
Given,
speed of conveyor belt, v = 3.2 m/s
coefficient of friction,f = 0.28
Using newton second law
f = ma
and we also know that frictional force
f = μ N
f = μ m g
equating both the force equation
a = μ g
a = 0.28 x 9.81
a = 2.75 m/s²
Using Kinematic equation
v = u + at
3.2 = 0 + 2.75 x t
t = 1.16 s.
Time taken by the box to move without slipping is 1.16 s.
Ответ:
I would say the answer is 3 because by falling technically the ball would be kind of moving in the air. Plus potential energy is when for example a soccer ball isnt moving