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3 years ago

6

You apply 100.0 N of force to a machine. The machine applies 20.0 N of force to the load. What is the mechanical advantage of th

e machine?

2 answers:

allsm [11]3 years ago

6 0

This machine has a mechanical advantage of 1/5, 20%, or no realistic advantage. This is because the machine transferred 80 N into external systems of the original 100 N applied.

Hope this helps!

max2010maxim [7]3 years ago

6 0

Answer:

0.20

Explanation:

The mechanical advantage of a machine is given by:

$MA=\frac{F_{out}}{F_{in}}$

where

Fout is the output force (the load)

Fin is the input force (the effort)

For this machine, the input force is 100.0 N, while the output force is 20.0 N, so the mechanical advantage is

$MA=\frac{20.0 N}{100.0 N}=0.20$

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2 years ago

As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, b

katovenus [111]

The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

<h3>What do you mean by sound radiates?</h3>

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(Lw) = 10·log (W/Wo) dB

Given:

sound level, $\beta= 101 dB$

Area, A = $22\;m^{2}$

Time, $\triangle t = 20\;min=1200\;s$

Intensity, $I=1\times 10^{-12}\;W/m^{2}$

$r=1\;km=1000\;m$

(a)

We know that, Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Solving the above equation for sound intensity,

$I=I_{o} \times 10^{\frac{\beta}{10} }$

$I=1 \times 10^{-12} \times 10^{\frac{101}{10} }$

$I=0.0126\;W/m^{2}$

Therefore, The sound energy is,

$E=P\times \triangle t$

Substitute $P=I \times A$ in the above equation,

$E=I \times A \times \triangle t$

$E=0.0126 \times 22 \times 1200$

$E=332.6\;J$

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

$A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi$

Substitute the known value in the above equation ,

$A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi$

$A_{hemisphere} = 6283185\;m^{2}$

Sound Intensity is,

$I = \frac{P}{A_{hemisphere}}$

Substitute $P=I \times A$ in the above equation,

$I = \frac{I \times A}{A_{hemisphere}}$

Substitute the known value in the above equation,

$I = \frac{0.0126 \times 22}{6283185}$

$I = 4.4 \times 10^{-8}\;W/m^{2}$

Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Substitute the known value in the above equation,

$\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )$

$\beta=46.4\;dB$

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Answer:

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Explanation:

According to the question

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3 years ago

A regular atom has a net charge of _____

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