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Cerrena [4.2K]
3 years ago
5

According to meet one law of universal gravitation, gravity increases when?​

Physics
1 answer:
mina [271]3 years ago
5 0

Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.

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Answer:

C the baseball began at rest and rolls at a rate of 14.7 m/s after 1.5 seconds

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Anaerobic transmission is when you touch a contaminated surface true or false
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2 years ago
What will be the restoring force if a spring with a spring constant of 45 newtons per meter is pulled 0.03 meters in the downwar
allsm [11]
To calculate for the force in a spring, we use Hooke's Law which relates force and the displacement of the spring. It is said that the force is directly proportional to the displacement. So, it will have the equation F = kx where k is a constant and it is the spring constant.

F = kx
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8 0
3 years ago
In which object more force has to be applied to move in between two objects of mas 11
jeka94

Answer:

my butt

Explanation:

2x2 by the power of 10 divide that 98

4 0
3 years ago
To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1
dedylja [7]

Answer:

2.2\times 10^{-5} s

Explanation:

We are given that  

The wavelength of sound wave=\lambda=3.3 cm=3.3\times 10^{-2}m/s

1 cm/s=10^{-2}m/s

Speed of sound wave,v=1522 m/s

We have to find the period of the wave.

We know that

Frequency=\nu=\frac{v}{\lambda}

Using the formula

Frequency =\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4} Hz

Time period=\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s

Using identity:\frac{a^x}{a^y}=a^{x-y}

Hence, the time period of the wave=2.2\times 10^{-5} s

4 0
3 years ago
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