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neonofarm [45]
3 years ago
10

The aluminum alloy (2024-t6) absorber plate is 6 mm thick and well insulated on its bottom. the top surface of the plate is sepa

rated from a transparent cover plate by an evacuated space. the tubes are spaced a distance / of 0.20 m from each other, and water is circulated through the tubes to remove the collected energy. the water may be assumed to be at a uniform temperature of 7z 60c. under steady-state operating conditions for which the qhw radiation heat flux to the surface is trad 800 w/m2 , what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? note that trad represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. you may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.
Physics
1 answer:
Xelga [282]3 years ago
5 0
The aluminum alloy(2024-T6; k=180 W/m<span>!K,) </span>absorber plate<span> is </span>6 mm thick<span> and </span>well insulated<span> on </span>its bottom<span>. The </span>top surface<span> of the </span>plate<span> is </span>separated<span> from a </span>transparent cover plate<span> by an </span>evacuated space<span>. The </span>tubes<span> are</span>spaced<span> a </span>distance<span> L of ...</span>
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The system needs an ordinary friction-based brake to bring the train to a full stop. Explain why the magnetic brake is not very
BabaBlast [244]

Answer:

The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.

Explanation:

A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.

Let us find the drag force which is proportional to the velocity of two conducting plates.

The EMF that is induced in the eddy currents are : $E=v(B \times L)$

The force which is due to the induced magnetic field is, $F=l(L \times B)$

Therefore, $F=\frac{E}{R} \times (L \times B)$

                 $F=\frac{v(B \times L)}{R} \times (L \times B)$

Here, force is directly proportional to the velocity of the two conducting plates.

Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.  

6 0
3 years ago
If there is 10% sales tax on a car costing $18,000
MaRussiya [10]

Answer:

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Explanation:

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6 0
3 years ago
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geniusboy [140]
I would say the same thing as the first answer
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4 years ago
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The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto
igor_vitrenko [27]

a) The angular velocity of the electron is 4.12\cdot 10^{16} rad/s

b) The number of revolutions per second is 6.54\cdot 10^{15} rev/s

c) The centripetal acceleration of the electron is 8.98\cdot 10^{22} m/s^2

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

\omega = \frac{v}{r}

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the angular velocity is

\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s

b)

The period of revolution of the electron is given by

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

where

\omega = 4.12\cdot 10^{16}rad/s is the angular velocity

Substituting,

T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15} rev/s

c)

The centripetal acceleration of an object in circular motion is

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circle

For the electron, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the centripetal acceleration is

a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0
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Low-pressure areas tend to ____ before occluding and ____ after occluding
alexgriva [62]

Out of the following given choices;

a. decelerate; accelerate

b. accelerate; decelerate

c. stop; accelerate

d. decelerate; stop

<span>The answer is B.  This is because the cold front usually rotates around the warm front as cold air mass coverage in the low-pressure system. This causes the cold air mass to accelerate to catch up with the warm front. When an occluded front ( that is the boundary that separates the older cool air mass already in place from new incoming cold air mass), that the system decelerates. </span>






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