The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by
Q(t)=t3−2t2+4t+4

Find the current when t = 1s. At what time is the current lowest?

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

At t = 1 s,

Current,

So, the current at t = 1 s is 3 A.

For lowest current,

Hence, this is the required solution.

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

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