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Wewaii [24]
3 years ago
11

The water in a deep underground well is used as the cold resevoir of a Carnot heat pump that maintains the temperature of a hous

e at 308 K. To deposit 13900 J of heat in the house, the heat pump requires 900 J of work. Determine the temperature of the well water.
Physics
1 answer:
VMariaS [17]3 years ago
6 0

Answer:

The temperature of well water be 288.07 K

Explanation:

We have given Heat in house Q_H=13900J

Work done W = 900 J

Efficiency is given by \eta =\frac{W}{Q_H}=\frac{900}{13900}=0.0647

Efficiency is also given by \eta =1-\frac{T_L}{T_H}, here T_L is lower temperature and T_H is higher temperature

So 0.0647=1-\frac{T_L}{T_H}

0.0647=1-\frac{T_L}{308}

T_L=288.07K

So the temperature of well water be 288.07 K

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Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m. Which statement a
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Answer:

a) Not Accurate

b) Not Accurate

c) Accurate

d) Accurate

Explanation:

Part a

Not Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m

Part b

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Part c

Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m by varying the phase difference between two waves she can achieve the desired results.

Part d  

Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m by varying the phase difference between two waves she can achieve the desired results.

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3 years ago
A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

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Explanation:

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. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

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3 years ago
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alukav5142 [94]
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Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

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1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3

v_1=-2.5\ m/s

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