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Olegator [25]
2 years ago
12

Please answer and ill give brainliest

Physics
1 answer:
11111nata11111 [884]2 years ago
4 0

I think least electricity is used between probably 7-8a.m. and 4-5p.m.

This is because, around those times, the suns already out. Depending on how sunny it is, it may not be as cold as all the other times of the day. And by then, buildings are typically already warmed all up. Everybody's body heat also may play a factor in buildings. ( if there is a ton of people )

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Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe
NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B )  Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0
3 years ago
Read 2 more answers
What is the degree of overall satisfication a person gets from life called?​
Alexus [3.1K]

Answer:

I remember learning about this in health class. I believe the answer is quality of life.

Explanation:

5 0
3 years ago
Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected
V125BC [204]
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
​
q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
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)(1.60×10
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C)
​

=5.51×10
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making (e) the best choice for this question.
3 0
3 years ago
The total consumption of electrical energy in the United States is about 1.0 1019 joules per year.
dexar [7]

Answer:

a) 316880878140.2895 Joules

b) 1056.26 Joules

c) 372.80103310622 km²

Explanation:

Energy consumption

E=\frac{P}{t}\\\Rightarrow E=\frac{1\times 10^{19}}{365.25\times 24\times 60\times 60}\\\Rightarrow E=316880878140.2895\ W

Average rate of electrical energy consumption is 316880878140.2895 Joules

Total population = 300 million

E=\frac{316880878140.2895}{300\times 10^6}\\\Rightarrow E=1056.26\ J

Each person consumes 1056.26 Joules.

Efficiency = 0.85

Power from Sun

0.85\times 1000=850\ W/m^2

Area

\frac{316880878140.2895}{850}=372801033.10622\ m^2

Converting to km²

372801033.10622=\frac{372801033.10622}{1000\times 1000}=372.80103310622\ km^2

Area required to collect the electrical energy used in the United States is 372.80103310622 km²

4 0
3 years ago
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anygoal [31]

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