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:
Force(Romeo moving) = 5,000 N
Explanation:
Given:
Mass of horse = 900 kg
Acceleration = 20 km/hr
Find:
Force(Romeo moving)
Computation:
Acceleration = 20 km/hr
Acceleration in m/s = 20 / 3.6 = 5.555556 m/s²
Force = m x a
Force(Romeo moving) = 900 x 5.555556
Force(Romeo moving) = 5,000 N
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Hi there! Lets see!
- m is mass, and its units are kg
- k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²
Therefore:
![\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7Bm%7D%7Bk%7D%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5B%5Cdfrac%7Bkg%7D%7Bs%5E2%7D%5D%7D%7D%20%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5Bkg%5D%7D%5Ccdot%20s%5E2%7D%20%3D%20%5Csqrt%7B%5Bs%5D%5E2%7D%20%3D%20s)
The period is given in seconds so the formula is dimensionally correct.