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IRISSAK [1]
4 years ago
11

A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of and performs 4.1 kJ of net work, and rejec

ts of heat, in a single cycle. What is thermal efficiency of a Carnot heat engine, operating between the same heat reservoirs?
37%

43%

50%

62%

56%
Physics
1 answer:
Sati [7]4 years ago
6 0
There are some missing data in the problem. The full text is the following:
"<span>A </span>real<span> (</span>non-Carnot<span>) </span>heat engine<span>, </span>operating between heat reservoirs<span> at </span>temperatures<span> of 710 K and 270 K </span>performs 4.1 kJ<span> of </span>net work<span>, and </span>rejects<span> 9.7 </span>kJ<span> of </span>heat<span>, in a </span>single cycle<span>. The </span>thermal efficiency<span> of a </span>Carnot heat<span> engine, operating between the same </span>heat<span> reservoirs, in percent, is closest to.."

Solution:
The efficiency of a Carnot cycle working between cold temperature </span>T_C and  hot temperature T_H is given by
\eta = 1 - \frac{T_C}{T_H}
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem, T_C=270 K and T_H=710 K, the efficiency is
\eta = 1 - \frac{270 K}{710 K}=0.62 = 62%

Therefore, the correct answer is D) 62 %.
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elena55 [62]

<em>The sensitivity of a mercury thermometer can be increased by using a smaller mercury bulb, thinner wall and smaller bore. A smaller bulb contains less mercury and hence absorbs heat in shorter time. As a result it can response faster to temperature change.</em>

8 0
3 years ago
Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv
Anuta_ua [19.1K]
This is a good time to review Newton's 3rd law of motion:
"For every action, there is an equal and opposite reaction."

Gravitational force always acts in pairs.
Whatever force the Earth attracts something with,
the thing attracts the Earth with exactly the same force.

If Earth attracts a person with a gravitational force of <span><span>7.0 × 10² </span>newtons,
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.

Your weight on Earth is the same as the Earth's weight on you !
</span>
5 0
4 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

6 0
3 years ago
A small bubble rises from the bottom of a lake, where the temperature and pressure are 4°C and 3.0 atm, to the water's surface,
qwelly [4]

Answer:

7.13mL

Explanation:

P₁V₁T₁ = P₂V₂T₂

P₁ = 3atm , V₁ = 2.1 mL , T₁ = 273 + 4 =277K

P₂ = 0.95atm , V₂ = ? , T₂ = 273 + 25 =298K

V₂ = P₁V₁T₂ / P₂T₁

V₂ = (3atm)(2.1 mL )(298K) / (0.95atm)(277K)

V₂ = 7.13mL

6 0
4 years ago
Ben and Jerry went for a drive. They decided to stop for a snack and have some ice cream. How fast did they drive on their way t
raketka [301]

Answer: B) 30 km/h

Explanation:

The speed of Ben and Jerry on their way to the store is given by

v=\frac{d}{t}

where

d is the distance they covered

t is the time they took

From the graph, we see that Ben and Jerry travelled for a distance of d=60 km in a time of t=2 h, so their speed is

v=\frac{60 km}{2 h}=30 km/h

5 0
3 years ago
Read 2 more answers
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