<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>
A negative to a negative charge will make a neutral charge.
Answer:
625000 N/ m
Explanation:
m= 20 kg
v= 30 m/s
x= 12 cm
k = ?
Here when the mass when hits at spring its speed is
Vi= 30 m/s
Finally it comes to rest after compressing for 12 cm
i-e Vf = 0 m/s
Distance= S= 12 cm = 0.12 m
using
2aS= Vf2 - Vi2
==> 2a ×0.12 = o- 30 × 30
==> a = 900 ÷ 0.24 = 3750 m/sec2
Now we know;
F = ma
F= -Kx
==> ma= -kx
==> 20 × 3750 = -K × 0.12
==> k = 625000 N/ m
Answer:
d = 421.83 m
Explanation:
It is given that,
Height, h = 396.9 m
Horizontal speed, v = 46.87 m/s
We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,
Now d is the distance covered by the cannonball. So,
Hence, this is the required solution.
