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vodomira [7]
3 years ago
6

If Earth were the size of a grape, how big would the Moon be? How far away would the Moon be from the Earth? How large would the

Sun be? How far away would it be from grape-sized Earth?​
Physics
1 answer:
dangina [55]3 years ago
3 0

Answer:

The moon would be the size of a raspberry. It would be as far a like 4 or 5 inches. The sun would be as big as an cantaloupe.The sun would be as far as 10 or 11 inches.

Explanation:

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
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(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
3 years ago
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