Answer:
only thing I think of when I see that is 'Just Wondering'
Explanation:
Answer:
E = 58.7 V/m
Explanation:
As we know that flux linked with the coil is given as

here we have


now we have

now the induced EMF is rate of change in magnetic flux

now for induced electric field in the coil is linked with the EMF as





Answer:
The approximate change in entropy is -14.72 J/K.
Explanation:
Given that,
Temperature = 22°C
Internal energy 
Final temperature = 16°C
We need to calculate the approximate change in entropy
Using formula of the entropy

Where,
= internal energy
T = average temperature
Put the value in to the formula


Hence, The approximate change in entropy is -14.72 J/K.
Answer:middle
Explanation:
Because it will make the seasaw balanced