Answer:
Because he loves hid girlfriend more than, u. He's looking for love
Explanation:
<span>The proton differs from the electron in sign although they have the same value. Like the electron, a proton will gain 215 electron-volts of eV in Kinetic energy. So 1.602Ă—10^-19 J * 215 = 344.43 * 10^(-19) J.
But K. E. = mv^2 / 2, so v^2 = 2 * K.E/m. The mass of a proton is 1.673 * 10^-27 kg. So v = âš(2 * 344.43 * 10^(-19))/1.673Ă—10^-27 = 688.86 * 10^(-19)/1.673Ă—10^(-27) = 411.75 * 10^(-19-(-27)) = âš411.75 * 10^(8) = 202196.56
Also for the electron we have v^2 = 2 * K.E/m but here mass, m, = 9.109 * 10^-31 kg. So we have v = âš(2 * 344.43 * 10^(-19)) / 9.109 * 10^-31 = 688.86 * 10^(-19)/ 9.109 * 10^-31 = 75.624 * 10^(-19 - (-31)) = 75.624 * 10^(21) and v = 2.749 * 10^11</span>
Initial velocity (u) = 2 m/s
Acceleration (a) = 10 m/s^2
Time taken (t) = 4 s
Let the final velocity be v.
By using the equation,
v = u + at, we get
or, v = 2 + 10 × 4
or, v = 2 + 40
or, v = 42
The final velocity is 42 m/s.
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)
The charge chould be moved to infinity
