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trapecia [35]
4 years ago
7

The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3

x10^22 kg, find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon
Physics
1 answer:
nika2105 [10]4 years ago
3 0

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

F=\dfrac{Gm_em_m}{r^2}

It means F\propto \dfrac{1}{r^2}

So,

\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}

r₁ = 3.9x10⁸ km

r₂= 1.5x10⁸ km

So,

\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}

Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.

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Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert
True [87]

To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since it is given that 1 acre = 4,050 m^2, so to reach that value, 1st let us divide both sides by 7,744:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then we multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

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7 0
3 years ago
Write the following number in scientific notation 156.60
Liula [17]

Answer:

1.566 x 10^2

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8 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
Ugo [173]

Answer:

f_{e}  = 1.9 cm

Explanation:

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    M = M₀ m_{e}

Where M₀ is the magnification of the objective and  m_{e} is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       m_{e} = 25 /  f_{e}

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/f_{e}  

1) Let's look for the focal length of the eyepiece (faith)

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     M = 400X = -400

     f_{e}  = - 12 25 /0.40 (-400)

     f_{e}  = 1.875 cm

Let's approximate two significant figures

    f_{e}  = 1.9 cm

8 0
4 years ago
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5 0
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Which of the following is most closely related to an activated complex?
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Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

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