A refrigerator draws 4.5 A of current while operating on a 120-V power line. If the refrigerator runs 50% of the time and electric power costs $0.15/kWh, how much does it cost to run this refrigerator for a 30-day month

$ 29.16

Explanation:

The following data were obtained from the question:

Current (I) = 4.5 A

Voltage (V) = 120 V

Time (t) = 30 days

1 KWh = $ 0.15

Cost for 30 days =.?

From the question given above, we were told that the refrigerator runs 50% of the time.

Thus, the time of operation of the refrigerator will be:

Time (t) = 30 × 50% = 30 × 50/100

= 15 days.

Next, we shall convert 15 days to hours. This can be obtained as follow:

1 day = 24 h

15 days = 15 days × 24 h / 1 day

15 days = 360 h

Next, we shall determine the energy consumed in KWh This can be obtained as follow:

Current (I) = 4.5 A

Voltage (V) = 120 V

Time (t) = 360 h

Energy (E) =?

E = ivt

E = 4.5 × 120 × 360

E = 194400 Wh

Next, we shall convert 194400 Wh to KWh. This can be obtained as follow:

1000 Wh = 1 KWh

Therefore,

194400 Wh = 194400 Wh × 1 KWh/ 1000 Wh

194400 Wh = 194.4 KWh

Thus the energy consumed is 194.4 KWh

Finally, we shall determine the cost of energy consumed. This can be obtained as follow:

From the question given above,

1 KWh cost $ 0.15

Therefore, 194.4 KWh will cost = 194.4 KWh × $ 0.15 = $ 29.16

Therefore, the cost of running the refrigerator for 30 days is $ 29.16

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg