Answer: 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
The speed of a proton that has been accelerated from rest is equal to [tex]8.8 \times 10^5\;m/s[/tex].
Given the following data:
Scientific data:
To determine the speed of a proton that has been accelerated from rest, we would apply the work-energy theorem:
Work done in accelerating the proton = Kinetic energy the proton possesses.
[tex]qV_d = \frac{1}{2} MV^2[/tex]
Making V the subject of formula, we have:
[tex]2qV_d=MV^2\\\\V=\sqrt{\frac{2qV_d}{M} }[/tex]
Substituting the given parameters into the formula, we have;
[tex]V= \sqrt{\frac{2 \times 1.6 \times 10^{-19}\times 4000}{1.67 \times 10^{-27}} }\\\\ V= \sqrt{\frac{1.28 \times 10^{-15}}{1.67 \times 10^{-27}} }\\\\V=\sqrt{7.67 \times 10^{11}}[/tex]
Velocity, V = [tex]8.8 \times 10^5\;m/s[/tex]
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