What is the efficiency of an engine that puts out 250 J of work for every 500 J of heat put into it?

1 answer:

5 0

If you put 300 J of heat into an engine with an efficiency of 0.35, how much work can be done? 3. How much energy must be put into an engine with an efficiency of 0.6 if 270 J of work are required? 4. An engine with an efficiency of 0.425 uses 1200 J of energy. Find the amount of energy wasted by the engine. 5. Calculate the efficiency of an engine operating between temperatures of 258 K and 600 K 6. An engine runs with its exhaust (cold) reservoir at a temperature of 200 K. To what temperature should the input (hot) temperature be set if an efficiency of 0.8 is desired? 7. Complete the following table of temperatures. Fahrenheit Celsius Kelvin 213 15 98.6 75 408

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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo

BigorU [14]

Answer:

$\lambda = 6.25\times10^{-9}$ = 625 nm

Explanation:

We now that for

for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....

d= distance between the slits, λ= wavelength of incident ray

for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.

Given

d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m

applying formula

λ= (d*y)/(D*n)

putting values we get

$\lambda = \frac{1.19\times10^{-3}\times4.97\times10^{-2}}{9.47\times10}$

on solving we get

$\lambda = 6.25\times10^{-9}$ = 625 nm

8 0

3 years ago

Two people pull on the wagon each with a constant 20N force. Both people pull to the left. What is the Net Force on the wagon?

eduard

Answer:

I dont understand what you are trying to ask

Explanation:

5 0

2 years ago

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$