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Otrada [13]
2 years ago
13

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.
Physics
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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Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done
blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=\frac{F}{m}

Hence acceleration of the block is given by:

\Rightarrow a=\frac{25}{2}=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+\frac{1}{2}at^2

Substitute the values

\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2

\Rightarrow S=56.25 m

Now work done is given by:

 W=F.S

W = 25×56.25

W =1406.25 J

3 0
2 years ago
What is a good college i can attend if i have Low B's and high C's as grades
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Answer:

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Explanation:

5 0
3 years ago
A rock is thrown from a cliff and hits the ground five seconds later at a distance of 50 m from the cliff how high was a cliff
Iteru [2.4K]

Answer:

122.5m

Explanation:

d= 1/2 (9.8) (5) ^2

3 0
2 years ago
A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount
alexgriva [62]

Answer:

W = 0.012 J

Explanation:

For this exercise let's use Hooke's law to find the spring constant

         F = K Δx

         K = F / Δx

         K = 3 / (0.16 - 0.11)

         K = 60 N / m

Work is defined by

         W = F. x = F x cos θ

in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1

         W = ∫ F dx        

         W = k ∫ x dx

we integrate

         W = k x² / 2

          W = ½ k x²

let's calculate

         W = ½ 60 (0.19 -0.17)²

         W = 0.012 J

8 0
2 years ago
The specific heat of gold is 0.126 J/g°C. What mass of gold would change its temperature from 25.0 °C to 60.0 °C with the additi
fiasKO [112]
<span>The amount of heat energy needed to increase the temperature of a substance by </span>\Delta T<span> is given by:
</span>Q=m C_s \Delta T<span>
where m is the mass of the substance, Cs is its specific heat capacity and </span>\Delta T<span> is the increase in temperature of the substance.

In this problem, we have a certain mass m of gold, with specific heat capacity </span>C_s=0.126 J/g^{\circ}C<span>, to which we add Q=2825 J of energy. Its temperature increases by </span>\Delta T=60-25=35 ^{\circ}C<span>. Therefore, if we re-arrange the previous equation, we can find the mass of the block of gold:
</span>m= \frac{Q}{C_s \Delta T} = \frac{2825J}{0.126\cdot 35}} =641 g<span>
So, the correct answer is B.</span>
3 0
3 years ago
Read 2 more answers
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