Answer:
The kinetic energy is 
Explanation:
From the question we are told that
The radius of the orbit is 
The gravitational force is 
The kinetic energy of the satellite is mathematically represented as
where v is the speed of the satellite which is mathematically represented as
=> 
substituting this into the equation
Now the gravitational force of the planet is mathematically represented as
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
![KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5B%5Cfrac%7BGMm%7D%7Br%5E2%7D%5D%20%2A%20r)
=> 
substituting values
Answer:
m = 5.22 kg
Explanation:
The force acting on the bucket is 52.2 N.
We need to find the mass of the bucket.
The force acting on the bucket is given by :
F = mg
g is acceleration due to gravity
m is mass
So, the mass of the bucket is 5.22 kg.
Answer:
what is your question because I seem to not see a question in this question....
It a stage known as the "Main Sequence" stage. Stars achieve nuclear stability at this point and remain in this stage until the end of their lives.
Have a nice day!
Answer:
Proof in explanataion
Explanation:
The basic dimensions are as follows:
MASS = M
LENGTH = L
TIME = T
i)
Given equation is:
where,
H = height (meters)
u = speed (m/s)
g = acceleration due to gravity (m/s²)
Sin Ф = constant (no unit)
So there dimensions will be:
H = [L]
u = [LT⁻¹]
g = [LT⁻²]
Sin Ф = no dimension
Therefore,
![[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]](https://tex.z-dn.net/?f=%5BL%5D%20%3D%20%5Cfrac%7B%5BLT%5E%7B-1%7D%5D%5E2%7D%7B%5BLT%5E%7B-2%7D%5D%7D%5C%5C%5C%5C%5C%20%5BL%5D%20%3D%20%5BL%5E%7B%282-1%29%7DT%5E%7B%28-2%2B2%29%7D%5D)
<u>[L] = [L]</u>
Hence, the equation is proven to be homogenous.
ii)
where,
F = Force = Newton = kg.m/s² = [MLT⁻²]
G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²
G = [M⁻¹L³T⁻²]
m₁ = m₂ = mass = kg = [M]
r = distance = m = [L]
Therefore,
![[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\](https://tex.z-dn.net/?f=%5BMLT%5E%7B-2%7D%5D%20%3D%20%5Cfrac%7B%5BM%5E%7B-1%7DL%5E%7B3%7DT%5E%7B-2%7D%5D%5BM%5D%5BM%5D%7D%7B%5BL%5D%5E2%7D%5C%5C%5C%5C%5C%20%5BMLT%5E%7B-2%7D%5D%20%3D%20%5BM%5E%7B%28-1%2B1%2B1%29%7DL%5E%7B%283-2%29%7DT%5E%7B-2%7D%5D%5C%5C%5C%5C)
<u>[MLT⁻²] = [MLT⁻²]</u>
Hence, the equation is proven to be homogenous.
