Acurrent of 3.6-a flows through a conductor. calculate how much charge passed through and cross-section of the conductor in 25sec.

90 C

Explanation:

Electric current: This can be defined as the rate of flow of electric charge in a circuit. This can be expressed mathematically as,

I = dQ/dt

dQ = Idt

∫dQ = ∫Idt

Q = It Equation 1

Where Q = amount of charge, I = current, t = time.

Given: I = 3.6 A, t = 25 s.

Substituting into equation 1,

Q = 3.6(25)

Q = 90 C.

Hence the amount of charge passing through the cross section of the conductor = 90 C

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

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