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3 years ago

What is the maximum efficiency of a heat engine whose operating temperatures are 590 ?c and 325 ?c?

1 answer:

7 0

<span>The maximum possible efficiency, i.e the efficiency of a Carnot engine , is give by the ratio of the absolute temperatures of hot and cold reservoir.
η_max = 1 - (T_c/T_h)

For this engine:
η_max = 1 - [ (20 +273)K/(600 + 273)K ] = 0.66 = 66%

The actual efficiency of the engine is 30%, i.e.
η = 0.3 ∙ 0.664 = 0.20 = 20 %

On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir:
η = W/Q_h

So the heat required from hot reservoir is:
Q_h = W/η = 1000J / 0.20 = 5000J</span>

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Are colavent bonds more likely to be found in compounds contaning both materals and nonmetals or compounds contaning only nonmet

atroni [7]

Answer:

bonds between 2 alike atoms or most non-metals get a covalent bond when reacting to the ... It tells what kinds of atomas are in a compound and how many.

Hope this helped

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7 0

3 years ago

Read 2 more answers

A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.

alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Here,

k = Spring constant

m = Mass

Our values are given as,

$m = 35g = 35*10^{-3}kg$

$f = 1 Hz$

Rearranging to find the spring constant we have that,

$k = (2\pi f \sqrt{m})^2$

$k = 4\pi^2 f^2 m$

$k = (4) (\pi)^2 (1) (35*10^{-3})$

$k = 1.38N/m$

Therefore the spring constant is 1.38N/m

7 0

3 years ago

It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?

ankoles [38]

Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed
Vf = 27.5 m/s Final speed

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2 -24.0 ^2)
160kj = 4680 x m
convert kilo joules to jeoules 160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg

4 0

3 years ago

A 8.10×10^3 ‑kg car is travelling at 25.8 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 3.9

KengaRu [80]

Answer:

The magnitude of the average drag force is 2412.34 N.

Explanation:

Given that,

Mass of car $m=8.10\times10^{-3}\ kg$

Velocity v = 25.8 m/s

Distance $d= 3.90\times10^{2}$

Speed of car = 13.1 m/s

Height = 12.5 m

We need to calculate the magnitude of the average drag force

Using equation kinetic energy

$K.E_{i}=K.E_{f}+P.E+F_{d}$

$\dfrac{1}{2}mv_{i}^2=\dfrac{}{}mv_{f}^2+mgh+F\times d$

Where, $v_{i}$ = initial velocity

$v_{f}$ = final velocity

h = height

g = acceleration due to gravity

$F_{d}$ =drag force

m = mass of the car

d = distance

Put the value into the formula

$\dfrac{1}{2}\times8.10\times10^{3}\times25.8=\dfrac{1}{2}\times8.10\times10^{3}\times13.1+8.10\times10^{3}\times9.8\times12.5+F\times3.90\times10^{2}$

$F=\dfrac{\dfrac{1}{2}\times8.10\times10^{3}\times25.8-\dfrac{1}{2}\times8.10\times10^{3}\times13.1-8.10\times10^{3}\times9.8\times12.5}{3.90\times10^{2}}$

$F=-2412.34\ N$

$|F|=2412.34\ N$

Hence, The magnitude of the average drag force is 2412.34 N.

4 0

3 years ago

Which of the following is not a

marissa [1.9K]

Answer:

polarization

Explanation:

6 0

3 years ago