aleeWriting 2-Firs, Project MUSE . From Scent EngrwAs 0 7/07-001 onnor 3 willson

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aleeWriting 2-Firs, Project MUSE . From Scent EngrwAs 0 7/07-001 onnor 3 willson (Student section: 11515) 2181 PHYS 0174 Basic Physics for Science and Engineering, CRN 11436 L Messages Courses Help Course Contents » » HWA , proton roton (mass m 1.67-102 kg) is being accelerated along a straight line at 3.62-105 m/s' in a machine. If the proton has an initial speed of 1.54 10' m/s and travels 3.34 cm, what then b ed? 3x10-8m/s ubmit Answer Incorrect. Tries 1/20 Prexious Tries d what is then the increase in its kinetic energy? ubmit Answer Tries 0/20 Post Discussion 0 O Type here to search

Explanation / Answer

Given,

m = 1.67 x 10^-27 ; a = 3.62 x 10^15 m/s^2 ; u = 1.54 x 10^7 m/s ; s = 3.34 cm = 0.0334 m ;

a)Let v be the final speed

We know from eqn of motion

v^2 = u^2 + 2 a s

v = sqrt[(1.57 x 10^7)^2 + 2 x 3.62 x 10^15 x 0.0334] = 2.21 x 10^7 m/s

Hence, v = 2.21 x 10^7 m/s

b)increase in kinetic energy will be given by:

delta KE = Ke2 - Ke1

delta KE = 1/2 m v2^2 - 1/2 m v1^2

delta KE = 0.5 x 1.67 x 10^-27 [(2.21 x 10^7)^2 - (1.57 x 10^7)^2] = 2.02 x 10^-13 J

Hence, delta KE = 2.02 x 10^-13 J

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