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2 years ago

What is the drawback to using superconductors?

Physics

1 answer:

Fittoniya [83]2 years ago

7 0

Answer:

Option A

The cost of keeping the semiconductor below the critical temperature is unreasonable

Explanation:

First of all, we need to understand what superconductors are. Superconductors are special materials that conduct electrical current with almost zero resistance. This means that there is little or no need for a voltage source to be connected to them. As a matter of fact, once a superconductor is connected to a power supply, one can remove the power supply and the current will still flow.

However, most superconducts can only conduct at very low temperatures up to -200 degrees Celcius. This is because, at that temperature, their atoms and molecules are relatively settled, hence they pose little or no resistance to the flow of current.

This as you can guess is extremely difficult to do, as you will need a lot of effort to cool it to that temperature and maintain it.

This makes option a the answer:

The cost of keeping the semiconductor below the critical temperature is unreasonable.

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Which of the following is strenuous?

Arada [10]

Answer:

I think it c

Explanation:

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3 years ago

Read 2 more answers

A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increase

ololo11 [35]

As per FBD while its accelerating upwards

we can say that

$F_n - mg = ma$

here normal force is given as

$F_n = 1050 N$

$W = 840 N$

now mass is given as

$m(9.8) = 840$

$m = 85.7 kg$

now we will have

$1050 - 840 = 85.7 \times a$

$a = 2.45 m/s^2$

Now while accelerating downwards we can say by FBD

$mg - F_n = ma$

again plug in all values

$840 - 588 = 85.7 \times a$

$a = 2.94 m/s^2$

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3 years ago

Look at your data. With three foods, Flock v

wolverine [178]

Answer:the first one was x

the second one is y

Explanation:

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3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

5 0

3 years ago

Water flows over a section of Niagara Falls at the rate of 1.4 × 106 kg/s and falls 49.8 m. How much power is generated by the f

Ad libitum [116K]

Answer:

Power= 6.84×10⁸ W

Explanation:

Given Data

Niagara falls at rate of=1.4×10⁶ kg/s

falls=49.8 m

To find

Power Generated

Solution

Regarding this problem

GPE (gravitational potential energy) declines each second is given from that you will find much the kinetic energy of the falling water is increasing each second.

So power can be found by follow

Power= dE/dt = d/dt (mgh)

Power= gh dm/dt

Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m

Power= 6.84×10⁸ W

7 0

3 years ago