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Maru [420]
3 years ago
15

A wrench 20 cm long lies along the positive y-axis and grips a bolt at the origin. A force is applied in the direction 0, 4, −3

at the end of the wrench. Find the magnitude of the force needed to supply 150 N · m of torque to the bolt. (Round your answer to the nearest whole number)
Physics
1 answer:
Andreyy893 years ago
6 0

Answer:

1250 N

Explanation:

length of wrench = 20 cm = 0.2 m

Write the length of wrench in vector form

r = 0.2 j metre

Direction of force= \frac{0i+4j-3k}{\sqrt{0 + 16 + 9}}=\frac{0i+4j-3k}{5}

Let the magnitude of force is F

So, write the force in vector form

F = F  \frac{0i+4j-3k}{\sqrt{0 + 16 + 9}}=\frac{0i+4j-3k}{5}

torque, T = 150 Nm

Torque = T = r x F

150 i =0.2j\times F\frac{0i+4j-3k}{5}

F = 1250 N

Thus, the magnitude of force is 1250 N.

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The frequencies that the listener at P will hear a maximum intensity are: 85.75 Hz, 171.5 Hz, 257.25 Hz and 343 Hz.

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Δx can be re-written as: n×\frac{v}{f}

where;

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4.0 m = \frac{n*343}{f}

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Now;  n = 1, 2, 3, 4 ; the frequencies that the listener at P will hear the maximum intensity will be

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f = 1 × 85.75 Hz

f = 85.75 Hz

when n= 2

f = 2 × 85.75 Hz

f = 171.5 Hz

when n= 3

f = 3  × 85.75 Hz

f = 257.25 Hz

when n= 4

f = 4 × 85.75 Hz

f = 343 Hz

∴ the frequencies that the listener at P will hear the maximum intensity will be : 85.75 Hz, 171.5 Hz, 257.25 Hz and 343 Hz for n =1,2,3, and 4 respectively.

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