Ask question

Ask question

All categories

2 years ago

5

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

reservoir. during that time, what is the maximum amount of work wmax that the engine might have performed?

1 answer:

tino4ka555 [31]2 years ago

6 0

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

You might be interested in

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

$\lambda = 0.186 m$

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

$f' = f_0 (\frac{v + v_o}{v - v_s})$

here we know that

$f_0 = 1750 Hz$

$v_o = 15 m/s$

$v_s = 15 m/s$

now we will have

$f' = (1750)(\frac{340 + 15}{340 - 15})$

$f' = 1911.5 Hz$

Part b)

Apparent wavelength is given by the formula

$\lambda = \frac{v_{relative}}{f_{app}}$

here we will have

$\lambda = \frac{340 + 15}{1911.5}$

$\lambda = 0.186 m$

3 0

2 years ago

the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be

Nezavi [6.7K]

D:the electrons from being attracted to the grid instead of the anode

5 0

2 years ago

Read 2 more answers

A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?

mart [117]

We use the equation of motion,

$S= ut+\frac{1}{2}at^{2}$

Here, S is the height, u is initial velocity and a is acceleration.

Given, $S = 20 \ ft$ $S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft} = 6.096 \ m$

As acorn falls from tree, therefore we take the value of $a = 9.8 \ m/s^2$ and initial velocity $u = 0$ .

Substituting these values in equation of motion,

$6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s$

Thus, the time taken by the acorn to fall 20 feet ( 6.096 m ) is 1.12 s.

5 0

2 years ago

A dog travels north for 18 meters, east for 8 meters, South for 27 meters and then west for 8 meters. What is the distance the d

Afina-wow [57]

I'm really not sure if this is right but I'll try.
The distance that the dog traveled is probably all of the distances added up. I would guess that it's 67 meters in total.
The displacement is a little more tricky but you pretty much have to put a mental map in your head. Since East and West are both 8 meters, they cancel each other out. He travels more southern and that means the displacement is 9 meters south of his original location

7 0

2 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

2 years ago