8.8 miles. You divide 22.4 & 2.55.
Answer:
- a.

- b.

Explanation:
<h3>
a.</h3>
The equation for the voltage V of discharging capacitor in an RC circuit at time t is:

where
is the initial voltage, and
is the time constant.
For our problem, we know

and

So





This gives us

and this is the time constant.
<h3>
b.</h3>
At t = 18.8 s we got:



That would be 110/22, which is 5 hours
Answer:
1) The angle of deflection will be less than 45° ( C )
2) The angle of deflection will be greater than 45° but less than 90° ( E )
Explanation:
1) Assuming that the force applied has a direction which is perpendicular to the Earth's magnetic field
∴ Fearth > Fapplied hence the angle of deflection will be < 45°
2) when the Fearth < Fapplied
the angle of deflection will be : > 45° but < 90°