Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
Definitely 4 because it’s u for I hi ub u
The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is
1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,
The distance from the center of the square to one of the corners is 
The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.
2) 
![V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a](https://tex.z-dn.net/?f=V_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%5Csqrt2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B-2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%7D%5C%5CV_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%7D%20%28%5Cfrac%7B1%7D%7B%5Csqrt2%7D-1%29%5C%5CV_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%20%284%5Ctimes%2010%5E%7B-5%7D%29%28-0.29%29%5C%5CV_b%20%3D%20%28-%5Cfrac%7B2.9%5Ctimes10%5E%7B-6%7D%7D%7B%5Cpi%5Cepsilon_0%7D%29%5Btex%5D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E3%29%20The%20work%20done%20on%20q3%20by%20q1%20and%20q2%20is%20equal%20to%20the%20difference%20between%20%20energies.%20This%20is%20the%20work-energy%20theorem.%20So%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20U_b%20-%20U_a)
From 50km/h to 0km/h in 0.5s we need next acceleration:
First we convert km/h in m/s:
50km/h = 50*1000/3600=13.8888 m/s
a = v/t = 13.88888/0.5 = 27.77777 m/s^2
Now we use Newton's law:
F=m*a
F=1700*27.7777 = 47222N
Answer: They have different rigidities.
Explanation:
