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iogann1982 [59]
3 years ago
7

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

as an initial speed of 1.60 × 104 m/s and travels 3.90 cm, what then is (a) its speed
(b) the increase in its kinetic energy?
Physics
1 answer:
Veronika [31]3 years ago
6 0

Answer:

(A) Speed will be 44.18\times 10^4m/sec

(b) Change in kinetic energy = 1560\times 10^{-19}      

Explanation:

We have given mass of proton m=1.67\times 10^{-27}kg

Acceleration of the proton a=2.50\times 10^{12}m/sec^2

Initial velocity u = 1.60\times 10^4 m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

v^2=u^2+2as

v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039

v=44.18\times 10^4m/sec

(b) Initial kinetic energy KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2

Final kinetic energy KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2

So change in kinetic energy \Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J

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