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ArbitrLikvidat [17]
3 years ago
12

Please Help!

Physics
1 answer:
irga5000 [103]3 years ago
6 0

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction =  0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ  = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F =  18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

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0.44m/s

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wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

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Check attachment for solution

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An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm
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Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

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Using equation of motion

v = u+gt

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

t = \dfrac{20}{9.8}

t= 2.04\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

s=0+\dfrac{1}{2}\times9.8\times(2.04)^2

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A 53 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.36. The acceleration of gravity is 9.8 m
tino4ka555 [31]

Answer:

42.6 m

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mass of crate m = 53 kg

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acceleration due to gravity, g = 9.8 m/s^2

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f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8

f = 185.114 N

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