Answer:
The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth
Explanation:
The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.
The relation to determine Roche limit is:
....(1)
Here
is radius of planet and
are density of planet and moon respectively.
According to the problem,
Density of Earth,
= 5.5 g/cm³
Density of Moon,
= 3.34 g/cm³
Consider
be the radius of the Earth.
Substitute the suitable values in the equation (1).
![Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }](https://tex.z-dn.net/?f=Roche%5C%20limit%3D2.423%5Ctimes%20R_%7BE%7D%5Ctimes%5Csqrt%5B3%5D%7B%5Cfrac%7B5.5%20%7D%7B3.34%20%7D%20%7D)

Answer:
Force that holds atoms together in a metallic substance.
Explanation:
Hope this helps? C:
~Chiena
Answer:
<em>physical quantities that cannot be defined in terms of other quantities.</em>
<em>Examples</em><em>:</em><em> </em><em>Length - meter (m)Time - second (s)Amount of substance - mole (mole)Electric current - ampere (A)Temperature - kelvin (K)Luminous intensity - candela (cd)Mass - kilogram (kg)</em>
Answer:
20 m
Explanation:
We'll begin by calculating the kinetic energy of the mass. This can be obtained as follow:
Mass (m) = 10 kg
Velocity (v) = 20 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 10 × 20²
KE = 5 × 400
KE = 2000 J
Finally, we shall the height to which the mass must be located in order to have potential energy that is the same as the kinetic energy. This can be obtained as follow:
Mass (m) = 10 kg
Acceleration due to gravity (g) = 10 m/s²
Potential energy (PE) = Kinetic energy (KE) = 2000 J
Height (h) =..?
PE = mgh
2000 = 10 × 10 × h
2000 = 100 × h
Divide both side by 100
h = 2000 / 100
h = 20 m
Thus, the object must be located at a height of 20 m in order to have potential energy that is the same as the kinetic energy.