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zhuklara [117]
3 years ago
7

if the coolant in a refrigerator was not compressed back into a liquid after it flowed through the refrigerator what would happe

n

Physics
2 answers:
Olin [163]3 years ago
4 0

The system would stop cooling, and simply recycle warm vapor through the refrigeration loop.

Explanation:

From the Mollier diagram we see that one part of the “refrigeration” loop is not complete. The step from 1 to 2 in the diagram would not occur.
This may occur due to a lack of sufficient refrigerant in the system or a faulty compressor. In any case, this results in warm “coolant” vapor passing into the receiver without any condensation (Step 3-4 cannot happen).
Without any liquid coolant available, the cooling effect of expansion of the liquid through the evaporator is lost (step from 5-6), and the system will simply use energy (and generate more heat) while recirculating warm vapor.

Pavlova-9 [17]3 years ago
3 0

Less heat would be removed from the refrigerator.

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Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


6 0
3 years ago
A car rounds a flat curve and experiences a centripetal force directed toward the center of the curve and perpendicular to the d
Usimov [2.4K]
<h2>Answer:</h2>

If a car is rounding a flat curve, it experiences a centripetal force that pulls it towards the center of the circle it is rotating in.

Now,

The centripetal force can be balanced by the centrifugal force caused due to the acceleration of the body at the high speed which counters the centripetal force and in turn <u>prevents the car from slipping down the curve.</u>

So,

If the car doesn't hit the gas then the <em><u>car will fall down from the curve</u></em> as the Centripetal force will exceed the Centrifugal force of the car.

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3 0
3 years ago
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timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

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N+mgcos\theta = \frac{mv^2}{r}

Here,

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a_T = -16.10ft/s^2

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

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kotykmax [81]
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C mass I am pretty sure
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