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Minchanka [31]
3 years ago
15

7. If 8 million kg of water flows over Niagara Falls each second, calculate the power available at the bottom of the falls.

Physics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

The power will be "3.92×10⁹ Watts". A further explanation is given below.

Explanation:

The given values as per the question,

Rate,

= 8 million kg

Distance,

= 50 m

Gravity,

= 9.8 m/s²

As we know,

The power will be:

⇒ Power = Rate\times Distance\times  Gravity

On putting the values, we get

⇒             =  8\times 10^6\times 50\times 9.8

⇒             =3.92\times 10^9 \  Watts

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In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles
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A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp
Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

pV=nRT

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

\frac{pV}{T}=nR

For a fixed amount of gas, n is constant, so we can write

\frac{pV}{T}=const.

Therefore, for a gas which undergoes a transformation we have

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K

So we can solve the formula for p_2, the final pressure:

p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa

4 0
3 years ago
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