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dimulka [17.4K]

3 years ago

10

A refrigerator is used to remove 84 kj/min of heat from a tank. If the electric power consumed by the refrigerator is 1. 2 kw, w

hat is the cop of the refrigerator?.

1 answer:

zvonat [6]3 years ago

3 0

Answer:

84 kj/min = 1.4 kj/sec

Power Out / Power In = Heat Out / Heat In - Coefficient of Performance

1.4 kj/sec / 1.2 kj/sec = 1.17 = COP

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Because one side of the spoon is convex and the other side is concave, the spoon’s different sides reflect differently.

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A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted

Snezhnost [94]

We can solve the problem by using Snell's law, which states
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium
$\theta_i$ is the angle of incidence
$n_r$ is the refractive index of the second medium
$\theta_r$ is the angle of refraction

In our problem, $n_i=1.00$ (refractive index of air), $\theta_i = 28.0^{\circ}$ and $n_r=1.63$ (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_r = \frac{1.00}{1.63} \sin 28.0^{\circ} = 0.288$
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3 years ago

A current of 17.0 mA is maintained in a single circular loop of 2.00 \mathrm{~m} circumference. A magnetic field of 0.800T is di

poizon [28]

The magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$ .

We have current carrying single circular loop placed in a magnetic field

parallel to the plane of the loop.

We have to determine the Magnetic moment of the loop.

<h3>What is the formula to calculate the magnetic moment of loop?</h3>

The formula to calculate the magnetic moment of the loop is -

M = NIA

where -

N - Number of turns

I - Current in the loop

A - Area of the loop

According to the question, we have -

I = 17mA = 0.017A

Circumference (C) = 2 meters

B = 0.8 T

Now -

C = 2

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$\pi$ r = 1

r = $\frac{1}{\pi }$

r = 0.32

Using the formula -

M = NIA

M = $\pi$ NI $r^{2}$

M = 3.14 x 1 x 0.017 x 0.32 x 0.32

M = 3.14 x 4 x 0.017

M = 5.5 x $10^{-3}$ A $m^{2}$

Hence, the magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$

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