A family uses an electric frying pan with a power rating of 1.2 X 10^3 W. Although the pan is thermostatically controlled, its element was drawing power for 6.3 X 10^2 min in a period of one month. Calculate the electrical energy in kWh used by the pan during the month

378 KWh

Explanation:

We'll begin by converting 1.2×10³ W to KW. This can be obtained as follow:

10³ W = 1 KW

Therefore,

1.2×10³ W = 1.2×10³ W × 1 KW / 10³ W

1.2×10³ W = 1.2 KW

Next, we shall convert 6.3×10² mins to hours (h). This can be obtained as follow:

60 mins = 1 h

Therefore,

6.3×10² mins = 6.3×10² mins × 1 h / 60 mins

6.3×10² mins = 10.5 h

Finally, we shall determine the electrical energy in KWh used for 1 month (i.e 30 days). This can be obtained as follow:

Power (P) = 1.2 KW

Time (t) for 1 month (30 days) = 10.5 h × 30

= 315 h

Energy (E) =?

E = Pt

E = 1.2 × 315

E = 378 KWh

Thus, the electrical energy used for 1 month (i.e 30 days) is 378 KWh.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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