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GREYUIT [131]
3 years ago
13

How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water

Physics
1 answer:
alekssr [168]3 years ago
5 0

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

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Please I need with this
guajiro [1.7K]
I think its b im not sure im in 8th grade
8 0
3 years ago
How might you describe the mathematical procedure of finding the displacement when an object travels in two opposite directions?
levacccp [35]
Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.

Displacement = 2 km - 2km = 0

Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>
4 0
3 years ago
Someone please help me with these questions! (The ones in the picture) Please I am super confused!
lozanna [386]

Answer:

c-d

Explanation:

3 0
2 years ago
How can a soccer ball and a small tennis ball bouncing relates to newtons first law interita?
Anna35 [415]

Answer:

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7 0
3 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
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