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GrogVix [38]
3 years ago
11

A stunt man is being pulled, at a constant velocity, along a rough road by a cable attached to a truck. The cable is parallel to

the ground. The mass of the stunt man is 109 kg, and the coefficient of kinetic friction between the road and the stunt man is 0.870. Find the tension in the cable.

Physics
2 answers:
nordsb [41]3 years ago
3 0

Answer:

the tension in the cable= 929.33N

Explanation:

the tension in the cable is equivalent to the frictional force experienced between the road and the stunt man

coefficient of friction= μ=\frac{F_{r} }{N}

μ=0.870

N= Normal force =m × g

m= 109kg

g= acceleration due to gravity =9.8m/s²

F_{r}= μ × N

F_{r}= μ × m × g

F_{r}= 0.870 × 109 × 9.8

F_{r}=929.33N

The tension in the cable is 929.33N

yan [13]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
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Answer:

Answer:

A. - 0.017N. It acts to the left.

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The net coulombs force on the charge is

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F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

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The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

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F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

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