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MArishka [77]
2 years ago
12

A proposed ocean thermal-energy conversion (OTEC) system is a heat engine that would operate between warm water (16°C) at the oc

ean's surface and cooler water (6°C) 1,300 m below the surface. What is the maximum possible efficiency of the system? Enter your answer accurate to two decimal places.
Physics
1 answer:
LenaWriter [7]2 years ago
6 0

Answer:

3.46 %

Explanation:

We have given cooler temperature that is T_c = 6°C =273+6=279 K

And the warm temperature of water that is T_h=16^{\circ}C=16+273=289\ K

The maximum possible efficiency is given by \eta =1-\frac{T_c}{T_h}

So maximum efficiency \eta =1-\frac{T_c}{T_h}=1-\frac{279}{289}=0.0346=3.46 %

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Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

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5 0
3 years ago
A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr
MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

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Applying the formula we have

1/v +1/u= 1/f

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7 0
3 years ago
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Q = cmAT
castortr0y [4]

Answer: Q=3000 cal

Explanation:

We are given the following formula:

Q=m. c. \Delta T   (1)

Where:

Q=3000 cal is the amount of heat

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c=1 cal/g \°C  is the specific heat of water

\Delta T  is the variation in temperature, which in this case is  \Delta T=30\°C-20\°C=10\°C  

Rewriting equation (1) with the known values at the right side, we will prove the result is 3000 cal:

Q=(300g)(1 cal/g \°C)(10\°C)   (2)

Q=3000 cal   This is the result

8 0
2 years ago
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