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AVprozaik [17]
3 years ago
14

Explain your results: How can Earth’s gravity affect the water when the water isn’t actually touching the Earth? (2 pts) Do you

think the Moon’s gravity is affecting the water in the beaker? Why or why not? (2 pts) If you were on the Moon, do you think you would feel the pull of Earth’s gravity? Explain! (2 pts)
Physics
1 answer:
Levart [38]3 years ago
5 0

Answer:

Because the Earth has so much gravity, it can hold water, land, and life in it's atmosphere.

(Not sure what beaker you are talking about, so sorry) But I don't think the moon's gravity would have an effect on a beaker of water because the Earth's gravity is much more than the moon's.

I think you would be able to feel a little bit of Earth's gravity  on the moon because the Earth's gravity pulled the moon into orbit, therefore, gravity on Earth my have some effect on the moon.

hope this helps!

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A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)
Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0
2 years ago
The object will feel minimum force of gravity at the
Alika [10]

Answer:

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4 0
2 years ago
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never [62]

Answer:

A) The wave that travels through the rail reaches the microphone first.

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3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
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Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

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From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

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B. is not a validated bu experimentation

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