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Dafna1 [17]
3 years ago
9

How does mass affect the gravitational pull of the Sun, Earth, and the Moon?

Physics
2 answers:
Contact [7]3 years ago
6 0
The gravitational force between the Sun and the Earth is about 3.54x1022 N. This force keeps the Earth orbiting around the Sun. The gravitational force from the other planets does slightly affect the Earth's orbit, but the gravitational pull from the other planets and the Moon is still very small
madreJ [45]3 years ago
3 0
Larger mass creates a stronger pull
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Gelneren [198K]

Answer:

response

Explanation:

Acceleration is your changing Velocity. An object that is ACCELERATING is experiencing a change in velocity. usually positive. if an object such as a car reduces velocity, it is called deceleration

3 0
3 years ago
You can see your image in a shiny, flat surface because the light waves are bouncing off the surface back at you. This is an exa
Alexxx [7]
I think it would be Reflection because the light in the ray reflects off of any flat or shiny object.... Hope this helps ^-^....
4 0
3 years ago
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A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o
jonny [76]

Answer:

I don't know

Explanation:

4 0
3 years ago
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A miner finds a small mineral fragment with a volume of 5.74 cm^3 and a mass of 28.7 g. What is the density of that mineral frag
statuscvo [17]
P=M÷V
p=28.7÷5.74
p=5

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8 0
3 years ago
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Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many
Zielflug [23.3K]

Answer:

The 6.25 \times 10^{23} electrons are moving through the superconductor per second.

Explanation:

Given :

Current I = 1 \times 10^{5} A

Charge of electron e = 1.6 \times 10^{-19} C

Time t = 1 sec

From the formula of current,

Current is the number of charges flowing per unit time.

   I = \frac{ne}{t}

Where n = number of charges means in our case number of electrons

   n = \frac{It}{e}

   n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }

   n = 6.25 \times 10^{23}

Therefore, 6.25 \times 10^{23} electrons are moving through the superconductor per second.

5 0
3 years ago
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