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3 years ago

11

You're using a wedge to split a log. you are hitting the wedge with a large hammer to drive it into the log. it takes a force of

1900 n to push the wedge into the wood. if the wedge moves 0.23 m into the log, how much work have you done on the wedge?

2 answers:

KIM [24]3 years ago

6 0

The work performed on an object is the force multiplied by the distance it is moved, provided the movement is parallel to the force. Since that is the case here, we can get the work by W=Fd=1900N x 0.23m = 437J. This energy is used to split the wood.

RoseWind [281]3 years ago

5 0

Answer: 437 J

Work done can be defined as the product of force and displacement. It is the work required to move an object to certain distance using a certain force.

$work=force\times displacement$

It is given that:

$Force=1900 N$

$Displacement=0.23 m$

Then, $work=1900N\times 0.23m = 437 N-m=437J$

Hence, the work done on the wedge to drive it into log is 437 J

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<h2>Question</h2>

why freezer is made in the upper part of refrigrator

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A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

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<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

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(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

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