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2 years ago

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The decibel level of a jackhammer is 125 dB relative to the threshold of hearing. Determine the decibel level if three jackhamme

rs operate side by side.

1 answer:

Vsevolod [243]2 years ago

5 0

Answer:

130 dB

Explanation:

The equation for decibel level is given by:

$D=10log(\frac{I}{I_n} )\\\\Where\ D\ is\ the \ decibel\ level\ in\ dB, I\ is\ the\ intensity\ in \ W/m^2, \\I_n\ is\ threshold\ intensity\ to\ the\ human\ ear=1*10^{-12}W/m^2\\\\Given\ that\ D=125dB, hence:\\\\125=10log(\frac{I}{1*10^{-12}} )\\\\12.5=log(\frac{I}{1*10^{-12}} )\\\\I=3.2\ W/m^2$

The intensity for 1 jack hammer is 3.2 W/m², therefore for 3 jack hammers, the intensity = 3 * 3.2 = 9.6 W/m²

$D=10\ log(\frac{I}{I_n} )\\\\D=10*log(\frac{9.6}{1*10^{-12}} )\\\\D=130\ dB$

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$\frac{4}{\pi }$ x $\frac{54}{2^2}$ / $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$ => $\frac{54}{27}$ $\frac{1.3^2}{2^2}$

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2 years ago

Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp

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Answer: 500 Watts

Explanation:

Power $P$ is the speed with which work $W$ is done. Its unit is Watts ( $W$ ), being $1 W=\frac{1 Joule}{1 s}$ .

Power is mathematically expressed as:

$P=\frac{W}{t}$ (1)

Where $t$ is the time during which work $W$ is performed.

On the other hand, the Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path. It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter ( $1J=(1N)(1m)=Nm$ ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (2)

In this case, we have the following data:

$F=500 N$

$d=10 m$

$t=10 s$

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

$W=(500 N)(10 m)$ (3)

$W=5000 J$ (4)

Substituting (4) in (1):

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