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Archy [21]
3 years ago
11

Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the c

oefficient of friction in the hall?
Physics
1 answer:
levacccp [35]3 years ago
6 0

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

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Answer:

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Explanation:

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The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

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Exercise or Getting grader Stomata
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Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
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Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

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F=\frac{kq_1q_2}{r^2}

F=\frac{kQ\cdot Q}{r^2}

F=\frac{kQ^2}{r^2}

Now the magnitude of charge is 2Q and is at a distance of \frac{r}{2}

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A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What
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Answer:

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