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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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mrs. knight drops her briefcase. when the briefcase lands on the floor, what is the reaction force to the action force of the br
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A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 82 ◦ with the positive x axis. T
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Answer:

\theta = 211.7 degree

Explanation:

First displacement of the particle is given as

r_1 = 11 m at 82 degree with positive X axis

so we can say

\vec r_1 = 11 cos82 \hat i + 11 sin82 \hat j

\vec r_1 = 1.53\hat i + 10.9 \hat j

resultant displacement of the particle after second displacement is given as

r = 8.7 m at 135 degree with positive X axis

so we can say

r = 8.7 cos135\hat i + 8.7 sin135\hat j

r = -6.15 \hat i + 6.15 \hat j

now we know that

r = r_1 + r_2

now we have

r_2 = r - r_1

so we will have

r_2 = (-6.15 \hat i + 6.15 \hat j) - (1.53\hat i + 10.9 \hat j)

r_2 = -7.68 \hat i - 4.75 \hat j

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