All categories

2 years ago

What is the formula for calculating the efficiency of a heat engine?

Physics

1 answer:

Olegator [25]2 years ago

7 0

Explanation:

The efficiency of a heat engine is defined as the ratio of the work output to the heat input into the system.

It is highly impossible to get an engine that is 100% efficient.
A heat engine that is 100% efficient suggests that all the heat input into the system is used to do work completely.
There is no heat loss.

Efficiency = $\frac{work output}{heat input}$ x 100

Efficiency has no unit

Learn more:

Heat brainly.com/question/914750

#learnwithBrainly

You might be interested in

Which describes a high frequency wave?

SCORPION-xisa [38]

I think the correct answer from the choices listed above is option A. A high frequency wave is a wave with a low level of energy and a high pitch. Frequency is the number of waves passing per second of time. Hope this answers the question.

8 0

3 years ago

What are the characteristics of high energy waves?

sladkih [1.3K]

The relationship between wavelength, frequency and energy of Electromagnetic Radiation is given by E = hf = hc/lamba -------(1) So from (1) there's a linear relationship between E and f. The higher the frequency, f, the higher the energy E. Also from (1) it is obvious that the lower the wavelength, lambda, the higher the energy, E. This means the answer is D.

4 0

2 years ago

Read 2 more answers

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago

What must always be included on the graph

Nookie1986 [14]

Clearly visible data points and appropriate labels on each access that include units

4 0

3 years ago

1. It s defficult to lift a heavier stone than the lighter one. why

stiks02 [169]

Answer:

lThe effect of the attraction of the earth on a bigger stone can be observed more than the effect of attraction of the earth on a smaller one. hence it is difficult to lift a large stone than the smaller one on the earth surface.

4 0

3 years ago