Answer:
They move outwards.
They don't intersect each other at any point.
They show the electric field.
Explanation:
Answer:
1.1648×10⁻¹¹ N
Explanation:
Using
F = qvBsinФ..................... Equation 1
Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.
Note: The angle between v and B = 90°
Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°
Substitute into equation 1
F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°
F = 11.648×10⁻¹²
F = 1.1648×10⁻¹¹ N.
Answer:
So percentage error will be 2 %
Explanation:
We have given initial value of acceleration due to gravity ![g=10m/sec^2](https://tex.z-dn.net/?f=g%3D10m%2Fsec%5E2)
And final value of acceleration due to gravity ![g=9.8m/sec^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fsec%5E2)
We have to find the percentage error
We know that percentage error is given by ![percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100](https://tex.z-dn.net/?f=percentage%5C%20error%3D%5Cfrac%7Binitial%5C%20value-final%5C%20value%7D%7Binitial%5C%20value%7D%5Ctimes%20100)
So
%
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.
For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left