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netineya [11]
2 years ago
5

What is the value of G on earth and moon?

Physics
2 answers:
skad [1K]2 years ago
8 0

We have that The acceleration due to gravity on the Moon And

The acceleration due to gravity on the surface of the Earth is approximately

1.625 m/s^2

9.81 m/s^2,

From the Question we are asked

The value of G on earth and moon?

Where G refers to The acceleration due to gravity

Generally

The acceleration due to gravity on the Moon is approximately

1.625 m/s^2,

And

The acceleration due to gravity on the surface of the Moon is approximately

9.81 m/s^2,

For more information on this visit

brainly.com/question/19927369

Nikolay [14]2 years ago
3 0

◆ The value of G on earth is 9.8m/s^2

◆ The value of G on the moon is 1.62m/s^2

I hope it's help you...

Thanks♥♥

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Boy X and Boy Y both move backward in opposite directions.
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The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
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Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

8 0
3 years ago
Read 2 more answers
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False
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This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa
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Answer:

2.77287\times 10^{15}\ m

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

\nu = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s

Photon intensity is given by

i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m

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Gauss's law: Group of answer choices can always be used to calculate the electric field. relates the electric field throughout s
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Answer:

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ε0 = the electric constant

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3 years ago
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