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Dovator [93]
3 years ago
12

A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the i

nput and output hold true. Which of the following sets of statements must apply for the heat engine to do work?
A) Qh < Qc and Th < Tc
B) Qh > Qc and Th < Tc
C) Qh < Qc and Th > Tc
D) Qh > Qc and Th > Tc
Physics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

Heat at the input must be greater than heat at the output for a heat engine to do work. Similarly the temperature at the input must be higher than the temperature at the output. Therefore, the right option is D.

Explanation:

Heat engines are designed to convert heat absorbed by a system to do work. Most times, there is an amount of heat lost from the system, thus, the amount of energy available to do work becomes lower than the heat gained.

<em>Q_{h} = W_{in} + Q_{c} ...... (1)</em>

Equation 1 shows that the heat at the input (hotter region) must be greater than heat at the output (colder region) for the system to do work.

The ratio of the work done by the engine to the heat gained from the hotter region is known as the <em>Coefficient of Performance (C.o.P).</em>

<em>C.o.P=\frac{W_{in} }{Q_{h}} ..... (2)</em>

Heat pumps and refrigerators are typical examples of heat engines. For heat pumps, the hotter region is the major concern that determines the performance of the engine, while the colder region determines the performance of the engine in refrigerators.

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____ [38]

Explanation:

formula for force is:

force=mass × acceleration

but in case of friction

force =coefficient of friction × Normal Reaction

F. = u × R

U = F/R

but when placed horizontally

R= M×G

M=mass=60kg

G=Gravity(10m/s or 9.8m/s)

F=140N

U=140/60×10

U=140/600

U=0.2333333333

approximately to 3 significant figures

U=0.233

if i am correct rate it 5 star

4 0
2 years ago
A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if
KengaRu [80]

Answer:

The value of  the correct angle of banking for the road is \theta = 67.76 °

Explanation:

Given data

Velocity (v) = 60 \frac{m}{s}

Radius = 150 m

The velocity of the car in this case is given by

v = \sqrt{r g \tan \theta}

v^{2} = r g \tan \theta

\tan \theta = \frac{v^{2} }{rg}

Put all the values in above formula we get

\tan \theta = \frac{60^{2} }{(150)(9.81)}

\tan \theta = 2.446

\theta = 67.76 °

Therefore the value of  the correct angle of banking for the road is \theta = 67.76 °

4 0
3 years ago
Fuel which has tremendous highest calorific value. (11 points if u answer this)
Vitek1552 [10]

Semi anthracite has the higest which is 29.5

8 0
3 years ago
Why do remote controls for TV’s use infrared waves to communicate?
Delvig [45]

Answer:  An IR remote (also called a transmitter) uses light to carry signals from the remote to the device so it can be controlled. It emits pulses of invisible infrared light that correspond to specific binary codes. These codes represent commands, such as power on, volume up, or channel down.

Explanation:

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3 years ago
A redecor travelling of 94 m/s s lows at a anstant
kotykmax [81]

Answer:

638 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 94 m/s

Final velocity (v) = 22 m/s

Time (t) = 11 s

Distance (s) =?

We can obtain the distance travelled by using the following formula:

s = (u + v) t /2

s = (94 + 22) × 11 /2

s = 116 × 11 /2

s = 1276 /2

s = 638 m

Thus, the distance travelled is 638 m.

3 0
3 years ago
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