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.
Series,effective resistance =R₁+R₂+R₃...
parallel,effective resistance 1/R=1/R₁ +1/R₂ +1/R₃...
Here,effective resistance 1/R =1/7 +1/5+ 1/4+1/2
=1.092
R = 1/1.092 =0.915Ω
voltage V=9 V
current I=V/R
I=9 / 0.915
=9.83 A
Answer:
The time is 133.5 sec.
Explanation:
Given that,
One side of cube = 10 cm
Intensity of electric field = 11 kV/m
Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.
We need to calculate the rate of energy transfer from the beam to the cube
Using formula of rate of energy


Put the value into the formula


We need to calculate the amount of heat
Using formula of heat


Put the value into the formula


We need to calculate the time
Using formula of time

Put the value into the formula


Hence, The time is 133.5 sec.
Answer:
Radius of the circle will be 2.5 m
Explanation:
We have given velocity of particle moving in the circle v = 5 m/sec
Acceleration of particle in the circle 
We have to find the radius of the circle
We know that acceleration is given by 
So 

So radius of the circle will be 2.5 m