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.

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west