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ivolga24 [154]
2 years ago
12

5. How much does a 20 m x 10 m x 8 m swimming pool filled with water weigh? Assume that water has a density of 62 kg/m².​

Physics
2 answers:
Licemer1 [7]2 years ago
6 0

Answer:

fxhehjejejkwkwkkwkwkqklwkjsnenrnkeke

scoray [572]2 years ago
3 0

Answer:

99,200kg

Explanation:

density= mass/volume

given d-62kg/m*3

v=20×10×8. =1600

m=x

1600×62=99,200kg

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak
Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

P = AI

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

A = \frac{\pi d^2}{4}

Now the intensity is inversely proportional to the square of the distance from the source, then

I \propto \frac{1}{r^2}

The expression for the intensity at different distance is

\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}

Here,

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I_2 = Intensity at distance 2

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r_2 = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

I_2 = I_1 (\frac{r_1^2}{r_2^2})

If we replace with our values at this equation we have,

I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})

I_2 = 1.11*10^{-4} W/m^2

Now using the equation to find the area we have that

A = \frac{\pi (8.4*10^{-3}m)^2}{4}

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Finally with the intensity and the area we can find the sound power, which is

P = AI

P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)

P = 6.1*10^{-9}J/s

Power is defined as the quantity of Energy per second, then

E = 6.1*10^{-9}J

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Leno4ka [110]

Answer:

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3 years ago
As additional resistors are connected in series to a constant voltage source, how is the power supplied by the source affected?
Furkat [3]

Answer:

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