A 9.5 V battery supplies a 2.5 mA current to a circuit for 7.0 h . Part A How much charge has been transferred from the negative to the positive terminal? Express your answer with the appropriate units. Δq Δ q = nothing nothing SubmitRequest Answer Part B How much electric potential energy has been gained by the charges that passed through the battery? Express your answer with the appropriate units. ΔU Δ U = nothing nothing SubmitRequest Answer Provide Feedback Next

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Answer:

A)Δq = 63 C

B)ΔU=595.5 J

Explanation:

Given that

V = 9.5 V

I = 2.5 mA

t= 7 h

A)

We know that

Charge = current X time

q= I X t

[tex]q=2.5\times 10^{-3}\times 7\times 60\times 60\ C[/tex]

q= 63 C

Δq = 63 C

B)

So the change in electric potential energy = I .V.t

Now by putting the values

[tex]Electric\ potential\ energy\ =2.5\times 10^{-3}\times 9.5\times 7\times 60\times 60\ J[/tex]

Electric potential energy=595.5 J

ΔU=595.5 J

Part A: The charge at the circuit is 63 C.

Part B: The change in the electric potential energy is 598.5 Joules.

How do you calculate the charge and potential energy?

Given that a 9.5 V battery supplies a 2.5 mA current to a circuit for 7.0 h.

Part A

The charge can be calculated as given below.

[tex]\Delta q = I \times t[/tex]

Where q is the charge, I is the current and t is time.

[tex]\Delta q = 2.5 \times 10^{-3}\times 7\times 60\times 60[/tex]

[tex]\Delta q = 63 \;\rm C[/tex]

Hence we can conclude that the charge at the circuit is 63 C.

Part B

The electric potential energy can be calculated as given below.

[tex]\Delta U = IVt[/tex]

Where U is the electric potential energy and V is the voltage across the circuit.

[tex]\Delta U= 2.5 \times 10^{-3} \times 9.5 \times 7\times 60\times 60[/tex]

[tex]\Delta U = 598.5 \;\rm J[/tex]

Hence we can conclude that the change in the electric potential energy is 598.5 Joules.

To know more about the charge and electric potential energy, follow the link given below.

https://brainly.com/question/9383604.