Question

How many watts does a flashlight that has 6.00×10^2 C pass through it in 0.500 h use if its voltage is 3.00 V?

a) 3.00 W
b) 36.0 W
c) 18.0 W
d) 9.00 W

Answer 1

The flashlight uses approximately 1.00 watt of power when a charge of 6.00×10² C passes through it in 0.500 hours with a voltage of 3.00 V.

Step-by-step explanation:

To determine how many watts the flashlight uses, we must first calculate the power (P) in watts (W).

The formula for power is P = IV, where 'I' is current in amperes (A) and 'V' is voltage in volts (V).

Given the charge (Q) of 6.00×10² C and the time (t) of 0.500 hours, we can first find the current by using the formula I = Q/t.

First, we convert 0.500 hours to seconds (0.500 h × 3600 s/h = 1800 s), since we need the time in seconds to get the current in the proper unit (amperes).

Now calculate the current:

I = Q/t

= 6.00×10² C / 1800 s

= 0.333 A

Then we calculate the power:

P = IV

= (0.333 A)(3.00 V)

= 0.999 W

The option closest to 0.999 W is (a) 1.00 W, hence the flashlight uses approximately 1.00 watt.