What is the speed of a proton that has been accelerated from rest through a potential difference of 4.0 kv?

v = 1.19×10^-8 m/s

Explanation: according to the work-energy theorem, the work done in accelerating the proton equals the kinetic energy that it gains.

Mathematically, we have that

qV = 1/2mv²

Where

q= magnitude of proton = 1.609×10^-16c

V = 4kv = 4000v

m = mass of proton = 9.11×10^-31 kg

v = velocity of proton =?

By substituting the parameters, we have that

1.609×10^-19 × 4000 = 1/2 × 9.11×10^-31 × v²

1.609×10^-19 × 4000 × 2 = 9.11×10^-31 × v²

v² = (1.609×10^-19 × 4000 × 2)/9.11×10^-31

v² = 12872×10^-19/9.11×10^-31

v² = 1412.95×10^-19

v² = 1.41×10^-16

v = √1.41×10^-16

v = 1.19×10^-8 m/s

i think the answer would be b

no sign of dia of fuel cell