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ddd [48]
3 years ago
7

How are size and mass of the moon different from that of the earth?

Physics
1 answer:
Elodia [21]3 years ago
3 0
<span>The diameter of the Moon is 3,474 km. Now, let's compare this to the Earth. The diameter of the Earth is 12,742 km. This means that the Moon is approximately 27% the size of the Earth.</span>
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Explain why an object’s weight is dependent upon where in the universe it is located.
strojnjashka [21]

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the  law of universal gravity

F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]

Now replacing the values in the equation we have:

F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a  radius of 71492 [km]. Find the force exerted by the planet upon the person.

F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} }  \\F= 1734.73[N]

We will calculate the value of the gravity accelaration of Jupiter. Let's check

g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

6 0
4 years ago
hii, just wondering how to solve this? it’s simple scientific notation but my brain just isn’t functioning today so it’s be help
arlik [135]
Honestly just use desmos or mathyway
8 0
2 years ago
Dimension of radius of sphere​
Readme [11.4K]

Answer:

The dimension is L

Explanation:

Dimension analysis is a method of representing quantities majorly with respect to some fundamental quantities of mass (M), length (L), time (T).

A sphere has a definite volume which relates to its radius by:

V = \frac{4}{3}\pir^{3}

In this equation \pi is a dimensionless quantity, and the unit of v is m^{3}.

But, metre is a measure of length, thus it has a dimension of L.

So that,

m^{3} ≅ L^{3}

Then,

L^{3} = r^{3}

Find the cube root of both sides to have,

r = L

Therefore, the dimension of the radius of a sphere is L.

5 0
3 years ago
Que es un kilogramo fuerza.?
Kipish [7]
El kilogramo de fuerza, o kilopond, es una unidad de fuerza métrica gravitacional. Es igual a la magnitud de la fuerza ejercida sobre un kilogramo de masa en un campo gravitatorio de 9.80665 m / s².
7 0
3 years ago
A magnetically soft material is placed in a strong magnetic field. What is the most likely outcome?
MakcuM [25]

Answer:

It will become a temporary magnet because the domains will easily realign.

Explanation:

3 0
3 years ago
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