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kupik [55]
4 years ago
8

A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the

coefficient of kinetic friction is 0.200. What is the friction force on the box if (a) a horizontal 140-N push is applied to it?
Physics
1 answer:
Dimas [21]4 years ago
3 0

Answer:

98N and 147N

Explanation:

We have the following information:

m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N

We can find the static fricton force as follow,

F=\mu_s * N

Where N is the normal force (mg)

F=0.3*50*9.8\\F=147N

Static friction force at 147N is greater than the force applied hence body does not move.

F=\mu_k N = 0.2*50*9.8= 98N

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Answer:

a) 800 km

b) 9.79° & 12.47°

Explanation:

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From the question, the time between the crest is given as 1.0 h, therefore, t = 1.0 h

Also, remember that frequency, f can be represented by

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入 = v/f, from the question, the speed is already stated as 800 km/h, so we apply directly

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B,

the smallest angle between the Southern tip of Africa is

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Sinθ = 0.17

θ = Sin^-1 0.17

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The smallest angle between the Southern end of Australia is

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Sinθ = 0.216

θ = Sin^-1 0.216

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The vertical component of a velocity vector points up and has a magnitude of 20, ms'. The velocity vector Itself makes an angle
kirill115 [55]

Answer:

Kindly check explanation

Explanation:

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Vertical component has Magnitude = 20m/s

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