SORRY i forgot but i think its A because the sun has a really strong gravitational pull
sorry and hope it helps
I believe the answer is B. that's where the asetroid belt is.
Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).
![U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}](https://tex.z-dn.net/?f=U_i%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%20%5Ctimes%20%5Cfrac%7B4%5Ctimes4q%5E2%7D%7B%5Csqrt%7B%2815%29%5E2%2B%285%2F%5Csqrt2%29%5E2%7D%7D)
It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:
![U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}](https://tex.z-dn.net/?f=U_f%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5C%20%5Ctimes%20%5Cfrac%7B4%5Ctimes4q%5E2%7D%7B%28%20%5Cfrac%7B5%7D%7B%5Csqrt%7B2%7D%7D%29%7D)
Potential energy shifts:
![= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\](https://tex.z-dn.net/?f=%3D%20U_f%20-U_i%20%5C%5C%5C%5C%20%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%5Csqrt2%7D%7B5%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7B%2815%29%5E2%2B%28%20%5Cfrac%7B5%7D%7B%5Csqrt%7B2%7D%29%5E2%29%7D%7D%5Cright%20%29%20%5C%5C%5C%5C%20%20%20%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%5Csqrt2%7D%7B5%7D-%5Cfrac%7B1%7D%7B%2015%20%2B%28%20%5Cfrac%7B5%7D%7B2%7D%29%7D%7D%5Cright%20%29%5C%5C%5C%5C%20%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%5Csqrt2%7D%7B5%7D-%5Cfrac%7B1%7D%7B%20%28%5Cfrac%7B30%2B5%7D%7B2%7D%29%7D%7D%5Cright%20%29%5C%5C%5C%5C)
![=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J](https://tex.z-dn.net/?f=%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%5Csqrt2%7D%7B5%7D-%5Cfrac%7B1%7D%7B%20%28%5Cfrac%7B35%7D%7B2%7D%29%7D%7D%5Cright%20%29%5C%5C%5C%5C%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%5Csqrt2%7D%7B5%7D-%5Cfrac%7B1%7D%7B17.5%7D%7D%5Cright%20%29%5C%5C%5C%5C%20%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%2024.74-%205%20%7D%7B87.5%7D%7D%5Cright%20%29%5C%5C%5C%5C%20%3D%5Cfrac%7B16q%5E2%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%20%5Cfrac%7B%2019.74-%205%20%7D%7B87.5%7D%7D%5Cright%20%29%5C%5C%5C%5C%20%3D%5Cfrac%7B4q%5E2%7D%7B%5Cpi%5Cepsilon_0%7D%5Cleft%20%28%200.2256%20%7D%5Cright%20%29%5C%5C%5C%5C%3D%20%5Cfrac%7B0.28%20%5Ctimes%20q%5E2%7D%7B%20%5Cepsilon_0%7D%5C%5C%5C%5C%3Dq%5E2%5Ctimes31.35%20%5Ctimes10%5E9%5C%2CJ)
Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.
![=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7BGMm%7D%7BR%7D-%5Cfrac%7BGMm%7D%7BR%2Bh%7D%20%5C%5C%5C%5C%3D%286.67%5Ctimes10%5E%7B-11%7D%5Ctimes6.0%5Ctimes10%5E%7B24%7D%5Ctimes100%29%5Cleft%28%5Cfrac%7B1%7D%7B6400%5Ctimes1000%7D-%5Cfrac%7B1%7D%7B6700%5Ctimes1000%7D%20%5Cright%20%29%20%5C%5C%5C%5C%20%3D%286.67%5Ctimes10%5E%7B-11%7D%5Ctimes6.0%5Ctimes10%5E%7B26%7D%29%5Cleft%28%5Cfrac%7B1%7D%7B64%5Ctimes10%5E%7B5%7D%7D-%5Cfrac%7B1%7D%7B67%5Ctimes10%5E%7B5%7D%7D%20%5Cright%20%29%20%5C%5C%5C%5C%3D%286.67%5Ctimes6.0%5Ctimes10%5E%7B15%7D%29%5Cleft%28%5Cfrac%7B67%20%5Ctimes%2010%5E%7B5%7D%20-%2064%20%5Ctimes%2010%5E%7B5%7D%20%20%7D%7B%204%2C228%20%5Ctimes10%5E%7B5%7D%7D%20%5Cright%20%29%20%5C%5C%5C%5C)
![=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\](https://tex.z-dn.net/?f=%3D%28%2040.02%5Ctimes10%5E%7B15%7D%29%5Cleft%28%5Cfrac%7B3%20%5Ctimes%2010%5E%7B5%7D%7D%7B%204%2C228%20%5Ctimes10%5E%7B5%7D%7D%20%5Cright%20%29%20%5C%5C%5C%5C%20%3D40.02%20%5Ctimes10%5E%7B15%7D%20%5Ctimes%200.0007%20%5C%5C%5C%5C)
![\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C](https://tex.z-dn.net/?f=%5C%5C%5C%5C%20%3D0.02799%5Ctimes10%5E%7B10%7D%5C%2CJ%20%5C%5C%5C%5C%3D%20q%5E2%5Ctimes31.35%5Ctimes10%5E%7B9%7D%20%5C%5C%5C%5C%20%3D0.02799%5Ctimes10%5E%7B10%7D%20%5C%5C%5C%5Cq%3D0.0945%5C%2CC)
This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Answer:
d d d d dd d d d d d dd d d d d dd d
Newton's 2nd law:
Fnet = ma
Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.
The electric force on a charged object is given by
Fe = Eq
Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.
We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe
Fe = Eq
Eq = ma
a = Eq/m
We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:
mass of proton = 1.67×10⁻²⁷kg
mass of electron = 9.11×10⁻³¹kg
The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.