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enot [183]
3 years ago
8

The mass of the moon is 7.36×1022kg and its distance to the Earth is 3.84×108m. What is the gravitational force of the moon on t

he earth?
Physics
1 answer:
marissa [1.9K]3 years ago
8 0

Answer:

The gravitational force is F =  2\,*\,10^{20}\,N

Explanation:

To answer this question we need to recall Newton's Universal Law of Gravitation for the force "F" exerted from one object to the other:

F=G\,\frac{m_1\,*\,m_2}{d^2}

where G is the Universal gravitational constant = 6.674\,* \,10^{-11}\,\,\frac{m^3}{kg\,s^2}

m_1, and m_2 are the masses of the two bodies/objects attracting each other via gravitational force. In our case, one is the mass of the Earth = 5.972\,*\,10^{24}\, kg

and the other one,the mass of the Moon = 7.36\,*\,10^{22}\,kg

and lastly, "d" is the distance between to two objects. In our case:

d =3.84\,*\,10^8\,m

Since all these quantities are given in SI units, when we use them in the formula, our answer will result in the SI units of force "N" (Newtons):

F=6.674\,*10^{-11}\,\frac{5.972\,10^{24}\,7.36\,10^{22}}{(3.84\,10^8)^2} \,N\\F=1.989\,*\,10^{20}\,N\\

which can be rounded to: F = 2\,*\,10^{20}\,N

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adell [148]

Answer:

7056 kJ

Explanation:

Given that,

Mass of a ship roller coaster is 36,000 kg.

It reaches a height of 20 m off the ground

We need to find the gravitational potential energy does it have. The formula for the gravitational potential energy ios given by :

E = mgh

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E = 36,000 kg × 9.8 m/s² × 20 m

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or

E = 7056 kJ

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3 years ago
To find the product of 183.5 and 10², move the decimal point in 183.5
S_A_V [24]

Answer:

Given :

To find the product of 183.5 and 10², move the decimal point in 183.5 _____ places to the right because 10² has _____ zeros.

To Find :

We need to fill the spaces.

Solution :

We know, when number with positive power of 10 is multiplied with another number we need to shift the number to the right hand side by the number of zeroes.

So, To find the product of 183.5 and 10², move the decimal point in 183.5 to right places to the right because 10² has two zeros.

Hence, this is the required solution.

Explanation:

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5 0
2 years ago
A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to s
Komok [63]
1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
2aS = v_f^2-v_i^2 = -v_i^2
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2

2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
F=ma
This force will be constant, and since m is always the same, then a is the same even in the second situation.

3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
2aS=-v_i^2
S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m
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What is the velocity of a wave with a frequency of 6 Hz and a wavelength of 2m?
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Frequency multiplied by the wavelength
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In the speed-time graph for a moving object shown here, the part which
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Answer:

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