Answer:
5. 9GmM/(10R)
Explanation:
m is the mass of the satellite
M is the mass of the earth
W is the energy required to launch the satellite
Energy at earth surface = Potential energy (PE) + W
W = Energy at earth surface - Potential energy (PE)
But PE = ![-\frac{GMm}{R}](https://tex.z-dn.net/?f=-%5Cfrac%7BGMm%7D%7BR%7D)
Therefore: W = Energy at earth surface - ![\frac{GMm}{R}](https://tex.z-dn.net/?f=%5Cfrac%7BGMm%7D%7BR%7D)
Energy at earth surface (E) at an altitude of 5R = ![-\frac{GMm}{5r} +\frac{1}{2}mV^2](https://tex.z-dn.net/?f=-%5Cfrac%7BGMm%7D%7B5r%7D%20%2B%5Cfrac%7B1%7D%7B2%7DmV%5E2)
But ![V=\sqrt{\frac{GM}{5R} }](https://tex.z-dn.net/?f=V%3D%5Csqrt%7B%5Cfrac%7BGM%7D%7B5R%7D%20%7D)
Therefore: ![E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}](https://tex.z-dn.net/?f=E%3D-%5Cfrac%7BGMm%7D%7B5R%7D%2B%5Cfrac%7B1%7D%7B2%7Dm%28%5Csqrt%7B%5Cfrac%7BGM%7D%7B5R%7D%20%7D%20%29%5E2%3D%20%20-%5Cfrac%7BGMm%7D%7B5R%7D%2B%5Cfrac%7BGMm%7D%7B10R%7D%20%20%3D%20-%5Cfrac%7BGMm%7D%7B10R%7D)
W = E - PE
![W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}](https://tex.z-dn.net/?f=W%3D-%5Cfrac%7BGMm%7D%7B10R%7D-%28-%5Cfrac%7BGMm%7D%7BR%7D%29%3D-%5Cfrac%7BGMm%7D%7B10R%7D%2B%5Cfrac%7BGMm%7D%7BR%7D%3D%5Cfrac%7B9GMm%7D%7B10R%7D%20%5C%5CW%3D%5Cfrac%7B9GMm%7D%7B10R%7D)
Answer:
How to find the maximum height of a projectile?
if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...
if α = 45°, then the equation may be written as: ...
if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.
Explanation:
Answer: I think Its the Height is 11.76 Meters (38.582677 Feet) between the bridge and the ground
Explanation: Supposing that where not counting air resistance in the equation, the equation
states that 1/2 multiplied by earths gravitational acceleration multiplied by the amount of time to reach the bottom: 2.4 seconds equals 11.76 meters of height between the bridge and the ground.
Magnets facing the same way <span />
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
![1 T : 4 A = t : 2 A](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%20t%20%3A%202%20A)
and solving for t we find
![t=\frac{(1T)(2 A)}{4A}=0.5 T](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%281T%29%282%20A%29%7D%7B4A%7D%3D0.5%20T)
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
![1 T : 4 A = t : 5 A](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%20t%20%3A%205%20A)
And by solving for t, we find
![t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%281T%29%285%20A%29%7D%7B4A%7D%3D%5Cfrac%7B5%7D%7B4%7D%20T%3D1.25%20T)