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3 years ago

14

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

n dry concrete where the coefficient of kinetic friction is 0.7 and the coefficient of static friction is 1.

1 answer:

Scilla [17]3 years ago

8 0

Answer:

Maximum acceleration will be equal to $3.43m/sec^2$

Explanation:

We have given coefficient of kinetic friction $\mu _k=0.7$

And coefficient of static friction $\mu _s=1$

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

When truck moves maximum force will be equal to $F=\mu _kmg$

It is given that half of the weight is supported by its drive wheels

So force required $=\frac{\mu _kmg}{2}$

From newtons law maximum acceleration will be equal to $a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2$

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A distance of 200 ft must be taped in a manner to ensure a standard deviation smaller than ± 0.04 ft. What must be the standard

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4 years ago

A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

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Answer:

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