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3 years ago

5

The cement bag shown in the figure has a mass of 50 kg and the small pulley has a mass of 2 kg. If the cement bag is pulled upwa

rds with an acceleration of 2 m/s^2, find the force F pulling the rope.

1 answer:

GuDViN [60]3 years ago

3 0

For the cement bag we can say as per its force diagram we will have

$2F - mg = ma$

here we will have

$m = 50 kg$

$a = 2 m/s^2$

now we will have

$F = \frac{mg + ma}{2}$

now plug in all data

$F = \frac{50(9.8 + 2)}{2}$

$F = 295 N$

so the pulling force will be 295 N

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3 years ago

A sound wave was determined to have a frequency of 0.3 Hz, speed of 150 cm/s, and amplitude of 2 cm. Find its wavelength.

fredd [130]

Answer:

5 m

Explanation:

From the question,

v = λf....................... Equation 1

Where v = speed of the sound wave, λ = wavelength of the sound wave, f = frequency of the sound wave.

make λ the subject of the equation

λ = v/f..................... Equation 2

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3 years ago

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

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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)

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

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

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

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

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

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(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:

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

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

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

(D)

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