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Lerok [7]
3 years ago
11

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

ge tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 5.90×10−2 kg . Throughout this problem use Lf=3.34×105J/kg for the heat of fusion for water. Part A During this time, how much work W is performed by the engin
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

W = 7.214 k J

Explanation:

given,

time of operation = 5 minutes

mass of ice melt = 5.90 × 10⁻² kg

L_f = 3.34 × 10⁵ J/kg

Work done = ?

We know work done

     |W| = |Q_h| - |Q_c|

Q_c = - m L_f

Q_c= - 5.90\times 10^{-2}\times 3.34 \times 10^5

Q_c = -19.706 kJ

Q_h = Q_c \dfrac{T_H}{T_C}

T_H is the heat from hot reservoir = 373 K

T_C is the energy released to cold reservoir = 273 K

Q_h = -19.706\times \dfrac{373}{273}

Q_h =-26.92 kJ

now,

     |W| = |-26.92| - |-19.706|

              W = 7.214 k J

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Inclined Plane

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2 years ago
A vector has components Ax = 12.0 m and Ay= 5.00 m. What is the angle that vector A makes with the x-axis?a. 67.4ob. 32.6oc. 22.
antoniya [11.8K]

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C.22.6°

Explanation:

Ax = ACosθ,

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Ax = 12.0 m,  Ay= 5.00 m

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3 years ago
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C

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6 0
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How tall would a tower need to be if the period of a pendulum were 30. seconds
Virty [35]
We use the following expression

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Where T is the period of the pendulum

l is the length of the pendulum

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4 0
3 years ago
A spherically-spreading EM wave comes from a 104.0 W source. At a distance of 9.6 m, what is the intensity of the wave?
Nookie1986 [14]

Answer:

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Explanation:

The intensity of light measures the power that the light delivers per unit area.

The source in this question delivers a constant power of \rm 104.0\; W. If the source here is a point source, that \rm 104.0\; W of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.

The surface area of a sphere of radius r is equal to 4\pi r^{2}. For the imaginary 9.6-meter sphere here, the surface area will be:

\rm 4\pi \times (9.6\; m)^{2} \approx 1158.12\; m^{2}.

That \rm 104.0\; W power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:

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8 0
3 years ago
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