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Hitman42 [59]
2 years ago
7

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, ho

w long will it take (in s) before the box moves without slipping
Physics
1 answer:
Lemur [1.5K]2 years ago
6 0

Answer:

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.

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A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV
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Answer:

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Explanation:

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6 0
3 years ago
A ball is thrown straight up with an initial speed of 12 m/s. What are the velocity and acceleration when it is at the top of it
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Answer:

'A ball is thrown straight up with an initial speed of 12 m/s_ What are the velocity and acceleration when it is at the top of its trajectory? Select all apply. v=12 mls a = 0 v =-12 mls a = 9.8 m/s2 Oa=-9.8 m/s2'

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