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rusak2 [61]
3 years ago
5

.

Physics
1 answer:
Greeley [361]3 years ago
7 0

Answer:

this is impossible for me

Explanation:

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Because the air in their wings helps them fly
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The specific latent heat of fusion of ice is 3.3×105J/kg
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Why is it important to use the correct number of significant digits when
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A. To show how precise your measurements were

Explanation:

It is important to use the correct number of significant digits when reporting the results of an experiment because they show how precise measurement are.

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A cylinder has a magnetic flux of 13 Wb directed inward through its bottom end, and a magnetic flux of 22 Wb directed out of the
satela [25.4K]

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Our data given are as follows:

Inward flux = 13Wb

Outward flux = 22Wb

Since the field of the second flow is directed OUT of the curved side we need to subtract the values, therefore,

Flux = 22-13

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4 0
3 years ago
If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy
Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

Given :

Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

Mass of each human, $m_p =65 \ kg$

Therefore mass of total population, $M_p = 65  \times 7 \times 10^{9} \ kg$

                                                           $M_p = 4.55 \times 10^{11} \ kg$

Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

         $M_m>> M_p$

Hence if we shift all the population on the moon there will be negligible change in the mass of the moon and earth. Hence there will not be any significant shift on the centre of mass. i.e.

      $X_{cm} = \frac{5.97 \times 10^{24}+ 7.34 \times 10^{22} \times 3.84 \times 10^5}{5.97 \times 10^{24}+ 7.34 \times 10^{22}}$

              $= 4.68 \times 10^6 \ m$

$ 4.68 \times 10^3 \ km$ from the earth.

         

3 0
3 years ago
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