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SCORPION-xisa [38]
2 years ago
13

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

ept at so that the engine efficiency is 0.79?
Physics
1 answer:
Paraphin [41]2 years ago
6 0
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
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Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
Which diagram shows how Rachel can see a candle flame?
SIZIF [17.4K]

Answer:

C

Explanation:

If the arrows represent light rays, then Rachel sees a candle flame when the light released by the flame is received by her eyes.

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2 years ago
Total resistance across any branch of a circuit can be found by analyzing whether the branch is connected in
atroni [7]

Answer: A.

series or parallel

Explanation:

Total resistance across any branch of a circuit can be found by analyzing whether the branch is connected in series or parallel.

The resistors are connected either in series or parallel. Therefore, the resistance of resistors across a circle can be calculated in series and parallel.

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Arrange the following substances from the lightest to the heaviest:
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molecular weights are written in the picture.

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A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de
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Answer: 7.734 m/s

Explanation:

We have the following data:

\theta=35\° The angle at which the water ballon was thrown

x=10 m  The horizontal distance of the water ballon

g=-9.8 m/s^{2} The acceleration due gravity

We need to find the initial velocity V_{o} at which the water ballon was thrown, and we can find it by the following equation:

x=V_{o}cos \theta T (1)

Where T=2t is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time T it takes the complete parabolic path.

So, if we use the following equation, we will find t:

V=V_{o}+gt=0 (2)

Isolating t:

t=\frac{-V_{o}}{g} (3)

Remembering T=2t:

T=2\frac{-V_{o}}{g} (4)

Substituting (4) in (1):

x=V_{o}cos \theta (2\frac{-V_{o}}{g}) (5)

Isolating V_{o}:

V_{o}=\sqrt{\frac{x g}{-2 cos \theta}} (6)

V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}} (7)

Finally:

V_{o}=7.734 m/s

4 0
2 years ago
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