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OlgaM077 [116]
3 years ago
14

A nuclear power plant generates a total of 150mw of power and 45% of this power is wasted heat that is discharged on the cooling

water of a river. the river flows at 660.43 gal/s and has an upstream temperature of 64.4 ⁰f. what is the temperature of the river downstream of the nuclear power plant (in k)?
Physics
1 answer:
Sedbober [7]3 years ago
8 0
The temperature of the river downstream of the nuclear power plant will be 53.3  degrees f

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What human practices are destroying the rain forest?
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Deforestation, the chopping off of the trees that can take thousands or even millions of years to grow again<span />
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3 years ago
An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit
mylen [45]

Find how much work ∆<em>W</em> is done by the motor in lifting the elevator:

<em>P</em> = ∆<em>W</em> / ∆<em>t</em>

where

• <em>P</em> = 45.0 kW = power provided by the motor

• ∆<em>W</em> = work done

• ∆<em>t</em> = 20.0 s = duration of time

Solve for ∆<em>W</em> :

∆<em>W</em> = <em>P</em> ∆<em>t</em> = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is <em>M</em> = (<em>m</em> + 490.0) kg, where <em>m</em> is the mass of the elevator alone. Then

∆<em>W</em> = <em>M</em> <em>g h</em>

where

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

• <em>h</em> = 35.0 m = distance covered by the elevator

Solve for <em>M</em>, then for <em>m</em> :

<em>M</em> = ∆<em>W</em> / (<em>g h</em>) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

<em>m</em> = <em>M</em> - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0
3 years ago
George saves 18% of his total gross weekly earnings from his two part-time jobs he earns $6.25 per hour from one part time job a
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Weekly wages and the amount saved each week

5 0
4 years ago
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A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag
ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B=10^{-7}\times \frac{2\times140}{5}

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0
3 years ago
A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0
3 years ago
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