The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed? 1. I = 0 always.

2. I = constant, not equal to 0.

3. I increases, then is constant.

4. I instantly jumps up, then slowly decreases.

5. None of the above.

The current in the ammeter is zero.

(1) is correct option.

Explanation:

Given that,

The capacitor is charged.

We need find the current after closed switched

We know that,

When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.

Then, the ammeter reads zero.

Hence, The current in the ammeter is zero.

(1) is correct option.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft