The shape is connected in parallel so;
5.1) Ans;

5.2) Ans;

I hope I helped you^_^
Answer:
The average magnetic flux through each turn of the inner solenoid is 
Explanation:
Given that,
Number of turns = 22 turns
Number of turns another coil = 330 turns
Length of solenoid = 21.0 cm
Diameter = 2.30 cm
Current in inner solenoid = 0.140 A
Rate = 1800 A/s
Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid
We need to calculate the magnetic flux
Using formula of magnetic flux


Put the value into the formula


Hence, The average magnetic flux through each turn of the inner solenoid is 
A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.
Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.
It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.
This question apparently comes after an EARLIER one,
where you were told either the voltage across the same
capacitor or the total charge stored in it. You can't answer
THIS one without that information.
Answer:
The force becomes 16 times what it is now.
Explanation:
The formula for gravitational force is
F = G * m1 * m2 / r^2
When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.
If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)
But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2
Because this is in the denominator, the 1/4 is going to flip to the numerator.
So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.
Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.