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Otrada [13]
3 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]3 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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Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

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3 years ago
Which transition represents a time when water molecules are moving closer together?
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. If she
I am Lyosha [343]

Answer:

9 meters

Explanation:

Given:

Mass of Avi is, m=40\ kg

Spring constant is, k=176,400\ N/m

Compression in the spring is, x=20\ cm=0.20\ m

Let the maximum height reached be 'h' m.

Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.

So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.

Decrease in elastic potential energy is given as:

EPE=\frac{1}{2}kx^2\\EPE=\frac{1}{2}\times 176400\times (0.20)^2\\EPE=88200\times 0.04=3528\ J

Now, increase in gravitational potential energy is given as:

GPE=mgh=40\times 9.8\times h=392h

Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,

392h=3528\\\\h=\frac{3528}{392}\\\\h=9\ m

Therefore, Avi will reach a maximum height of 9 meters.

6 0
3 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
3 years ago
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