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mixas84 [53]
3 years ago
14

In a recent year, wind machines in a country generated 183.1 billion kWh of electricity, enough to serve over 17 million househo

lds. The nation's total electricity production that year was 4,098 billion kWh. What percent of the total energy production was generated by wind machines
Physics
1 answer:
Neko [114]3 years ago
5 0

Answer:

The right answer is "4.4%".

Explanation:

Given:

Generated electricity,

= 183.1 billion kWh

Total electricity production,

= 4098 billion kWh

Now,

The wind energy percentage will be:

= \frac{183.1}{4098}\times 100

= 0.0446\times 100

= 4.4 %

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How do you use these muscles in your everyday life? What daily activities do you complete that mimic the movements of these exer
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While loading, a truck moves 10 meters west from point A to point B in 5 seconds. Then it moves back to point A in 5 seconds. Wh
Talja [164]

Answer:

2m/s

Explanation:

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time taken(t)=5s

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If a mineral breaks into cubes when struck with a rock hsmmer, which property is being measured? a.) hardness b.) cleavage c.) c
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8 0
3 years ago
Read 2 more answers
If the Hubble Space Telescope is 598 m above the surface of the Earth and is traveling at 7.56 x 103 m/s, how long does it take
viva [34]

Answer:

5069.04 seconds

Explanation:

The parameter we are looking for is called the Orbital period of the Hubble Space Telescope.

It is given as:

T = \sqrt{\frac{4\pi^2r^3 }{GM} }

where r = radius of orbit of Hubble Space Telescope

G = gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

M = Mass of earth

We are given that:

r = radius of the earth + distance of HST from earth

r = 6.38 * 10^6 + 598 = 6380598 m

M = 5.98 * 10^{24} kg

Therefore, T will be:

T = \sqrt{\frac{4*\pi^2 * 6380598^3 }{6.67408 * 10^{-11} * 5.98 * 10^{24}}}

T = 5069.04 secs

The orbital period of the Hubble Space Telescope is 5069.04 seconds.

7 0
3 years ago
Read 2 more answers
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