So, the force of gravity that the asteroid and the planet have on each other approximately ![\boxed{\sf{2.9 \times 10^{17} \: N}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B2.9%20%5Ctimes%2010%5E%7B17%7D%20%5C%3A%20N%7D%7D%20)
<h3>Introduction</h3>
Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :
![\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7BF%20%3D%20G%20%5Ctimes%20%5Cfrac%7Bm_1%20%5Ctimes%20m_2%7D%7Br%5E2%7D%7D%7D%7D%20)
With the following condition :
- F = gravitational force (N)
- G = gravity constant ≈
N.m²/kg²
= mass of the first object (kg)
= mass of the second object (kg)- r = distance between two objects (m)
<h3>Problem Solving</h3>
We know that :
- G = gravity constant ≈
N.m²/kg²
= mass of the planet X =
kg.
= mass of the planet Y =
kg.- r = distance between two objects =
m.
What was asked :
- F = gravitational force = ... N
Step by step :
![\sf{F = G \times \frac{m_X \times m_Y}{r^2}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%20G%20%5Ctimes%20%5Cfrac%7Bm_X%20%5Ctimes%20m_Y%7D%7Br%5E2%7D%7D%20)
![\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%3D%206.67%20%5Ccdot%2010%5E%7B-11%7D%20%5Ctimes%20%5Cfrac%7B1.55%20%5Ccdot%2010%5E%7B22%7D%20%5Ccdot%203.95%20%5Ctimes%2010%5E%7B28%7D%7D%7B%283.75%20%5Ctimes%2010%5E%7B11%7D%29%5E2%7D%7D%20)
![\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%5Capprox%20%5Cfrac%7B40.84%20%5Ctimes%2010%5E%7B-11%20%2B%2022%20%2B%2028%7D%7D%7B14.0625%20%5Ctimes%2010%5E%7B22%7D%7D%7D%20)
![\sf{F \approx 2.9 \times 10^{39 - 22}}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%5Capprox%202.9%20%5Ctimes%2010%5E%7B39%20-%2022%7D%7D%20)
![\sf{F \approx 2.9 \times 10^{17} \: N}](https://tex.z-dn.net/?f=%20%5Csf%7BF%20%5Capprox%202.9%20%5Ctimes%2010%5E%7B17%7D%20%5C%3A%20N%7D%20)
<h3>Conclusion</h3>
So, the force of gravity that the asteroid and the planet have on each other approximately
![\boxed{\sf{2.9 \times 10^{17} \: N}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B2.9%20%5Ctimes%2010%5E%7B17%7D%20%5C%3A%20N%7D%7D%20)
<h3>See More</h3>
<span>the deeper you go the more pressure...that's for depth</span>
Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
Answer:
Mass of bike = 38 kg.
Explanation:
Kinetic energy is given by the expression,
, where m is the mass and v is the velocity.
Here speed of child riding bike = 6 m/s
Mass of child = 30 kg
Total kinetic energy = 1224 J
Let the mass of bike be, m kg
So, total mass of child and bike = (m + 30) kg
Substituting,
![1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg](https://tex.z-dn.net/?f=1224%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%20%28m%2B30%29%2A6%5E2%5C%5C%20%5C%5C%20m%2B30%3D68%5C%5C%20%5C%5C%20m%3D38kg)
So, mass of bike = 38 kg.