Answer:
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.
Neutrons have a zero charge but consist of mass.
Answer:
I think the answer is a no. I guess
Explanation:
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Answer:
Answer if this question neon street sign
Explanation:
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Data Analysis and Conclusion
