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Sav [38]
3 years ago
12

A worker pushes horizontally on a 36.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the

crate and the floor is 0.380. (a) What is the value of fs,max under the circumstances
Physics
2 answers:
Yuliya22 [10]3 years ago
5 0

Answer:

value of fs,max under the circumstances is

f = 42.56 N

Explanation:

The formula is µ = f / N, where µ is the coefficient of friction, f is the amount of force that resists motion, and N is the normal force. Normal force is the force at which one surface is being pushed into another.

So f = µN

f = 0.380 * 112

f = 42.56 N

ICE Princess25 [194]3 years ago
4 0

The maximum value for the force of friction is given by:

F_{max}=\mu_sN

Where \mu_s is the static friction between the body and the surface and N is the normal force on the body. Using the free body diagram, we know that:

N=mg

Replacing this and solving:

F_{max}=\mu_s mg\\F_{max}=0.38*36kg*9.8\frac{m}{s^2}\\F_{max}=134.06N

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145 m

Explanation:

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substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

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6 0
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1500 kg Peugeot car is traveling at 16.67 m s ⁻¹ and accelerates to 30.56 m s ⁻¹ for 2 minutes. Calculate the impulse of the car
Elis [28]

Answer:

2084 kg*m/s

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Impulse is change in momentum

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