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2 years ago

Estimate humanity’s current consumption of energy in one day (in Joule).

1 answer:

6 0

Based on data from world energy consumption, the estimated humanity’s current consumption of energy in one day is above 208 million joules per day.

<h3>What is energy consumption?</h3>

Energy is the ability to do work. Therefore, energy consumption refers to the amount of energy used to do work.

Humans need energy for various purposes such as;

lighting
heating
operation of machines
feeding, etc.

Due to rapid growth on population as well as technology, the human energy consumption is on the increase.

In 2020, the world primary energy consumption was 71,4 GJ per person for a world population of about 7.7 billion people.

Therefore, the estimated humanity’s current consumption of energy in one day is above 208 million joules per day.

Learn more about energy at: brainly.com/question/6572692

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Air is composed of many gases, while water is composed of a pure substance. Which best compares air and water?

natta225 [31]

Answer: c

Explanation:

C Air is a compound of two or more components that keep their own identifying properties, while water is composed of mixtures that combine to form a compound.

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3 years ago

Read 2 more answers

If two connected points objects pass through the same set of three points, the shapes created by each will be identical, regardl

NeX [460]

Answer:

This is false

Explanation:

This is why the answer to this question is false. If these types of objects (2 points connected) should pass through same sets of 3 points, irrespective of the order that each object was plotted, we will not get identical shapes created.

The shape that is going to be created will be dependent on the pattern/order that was involved in the plotting. If it was identical, then we will have identical shapes. But if the order during plotting is different then we will have different shapes created.

Thank you!

5 0

3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$