As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.
The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.
The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.
A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.
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I think you almost got it.
At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.
With the value of the initial speed (28 m/s, which is the total speed), you can set
v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7
or theta = 64.62 deg, it is D. Think about it. I hope you see it.
