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Airida [17]
3 years ago
7

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe

fficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Physics
1 answer:
Levart [38]3 years ago
3 0

Answer:

Coefficient of static friction will be equal to 0.642  

Explanation:

We have given acceleration a=6.3m/sec^2

Acceleration due to gravity g=9.8m/sec^2

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to a=\mu g, here \mu is coefficient of static friction and g is acceleration due to gravity

So \mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642

So coefficient of static friction will be equal to 0.642

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3 years ago
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7 0
3 years ago
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Answer:

correct option is a. True

Explanation:

solution

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GenaCL600 [577]
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The vertebral region is _________ to the scapular region.
Lisa [10]

Answer:

<em>The answer is medial!</em>

Explanation:

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