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guajiro [1.7K]
3 years ago
8

20 points !!!

Physics
2 answers:
patriot [66]3 years ago
7 0
<span>A lot of things take up electricity. Just leaving something plugged in can cost energy that could have been used for something more worth while. I think the biggest thing people do that wastes energy is that we leave lights on when we don't need them. I would like to improve this method by turning my lights off everytime I exit the room or by not even turning on the light if I can see without it. If it is daylight outside and you have a window in the room and you can see perfectly fine without the light, then why turn it on? No one ever thinks about that. Leaing our lights on is a common factor that raises our light bills. We always think we are gonna come back into the room in a minute but it ends up being a hour later. That's valuable money down the drain because no one is in the room using the light.

</span>
beks73 [17]3 years ago
7 0

Answer:

A lot of things take up electricity. Just leaving something plugged in can cost energy that could have been used for something more worth while. I think the biggest thing people do that wastes energy is that we leave lights on when we don't need them. I would like to improve this method by turning my lights off everytime I exit the room or by not even turning on the light if I can see without it. If it is daylight outside and you have a window in the room and you can see perfectly fine without the light, then why turn it on? No one ever thinks about that. Leaing our lights on is a common factor that raises our light bills. We always think we are gonna come back into the room in a minute but it ends up being a hour later. That's valuable money down the drain because no one is in the room using the light.

Explanation:

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Assume that a resistor is connected between the 150 V terminal and the common terminal. The voltmeter is then connected to an un
Lena [83]

Answer: 316.8V

Explanation:

given data:

metre moving current = 0.96mA

meters voltage = 288V

or  0.96*300V = 288V

<u><em>Solution:</em></u>

<u><em /></u>v1 = (0.96mA*150)<u><em /></u>

<u><em /></u>= 144V<u><em /></u>

<u><em /></u>

i1 = \frac{144v}{750}

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