Answer:
3.99*10^-3N/C
Explanation:
Using
Ep= kq/r²
Where r = 0.6mm = 0.6*10^-3m
K= 8.9*10^9 and q= 1.6*10^-19
So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²
= 3.99*10^-3N/C
Answer:
Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Explanation:
Given;
initial speed of proton, u = 2.5 x 10⁵ m/s
initial potential, V = 1500 V
mass of proton = 1.67 x 10⁻²⁷ kg
Work done, W = eV= ΔK.E = ¹/₂mu²
eV = ¹/₂mu² (J)
where;
e is the charge of the proton in coulombs
V is the electric potential in volts
m is the mass of the proton in kg
u is the speed of the proton in m/s


Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
The answers is hydrochloride acids
The force that one massive object exerts to attract another object is called gravity or gravitational pull.
Move the fulcrum closer to the load