Answer to Question #106674 in Physics for grant

Question #106674
A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 6.10 × 1010 m/s2 in a machine. If the proton has an initial speed of 7.50 × 105 m/s and travels 1.50 cm, what then is (a) its speed and (b) the increase in its kinetic energy?
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Expert's answer
2020-03-30T07:54:18-0400

a)


v2=v02+2asv^2=v_0^2+2as

v2=(7.5105)2+2(6.11010)(0.015)v^2=(7.5\cdot10^5)^2+2(6.1\cdot10^{10})(0.015)

v=7.51105msv=7.51\cdot10^5\frac{m}{s}

b)


ΔK=0.5m(v2v02)=0.5m(2as)=mas\Delta K=0.5m(v^2-v_0^2)=0.5m(2as)=mas

ΔK=(1.671027)(6.11010)(0.015)=1.531018 J\Delta K=(1.67\cdot10^{-27})(6.1\cdot10^{10})(0.015)=1.53\cdot10^{-18}\ J


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