No Service 10:18 AM * 35% PRACTICE IT Use the worked example above to help you s

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No Service 10:18 AM * 35% PRACTICE IT Use the worked example above to help you solve this problem. A proton is released from rest at x =-2.00 cm in a constant electric field with magnitude 1.50 x 103 N/C pointing in the positive X-direction. (a) Assuming an initial speed of zero, find the speed of a proton at x = 0.0400 m with a potential energy of -1.44 x 10-J. (Assume the potential energy at the point of release is zero.) 1.31e+05 m/s (b) An electron is now fired in the same direction from the same position. Find the initial speed of the electron (at x =-2.00 cm) given that its speed has fallen by half when it reaches x = 0.110 m, a change in potential energy of 3.12 x 10-13. 9.56e+06m/s EXERCISE HINTS: GETTING STARTED I'M STUCK! The electron in part (b) travels from x = 0.110 m (where it has half the initial speed you previously calculated) to x-0.280 m within the constant electric field. If there's a change in electric potential energy of-9.37 × 10-17 J as it goes from x . 0.110 m to x·-0.280 m, find the electron's speed at x =-0.280 m. (Note: Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.) Yv=1 1.44e+07 |X What factors will affect the velocity of the electron? How far has the electron moved in this instance? Through what field is it moving? m/s Need Help?Read lit Submit Answer Save ProgressPractice Another Version 3. 1.05/1 points | Previous Answers SerCP11 16.QQ.001 My Notes Ask Your

Explanation / Answer

displacement, s = 0.39 m, u= -4.78*106 m/s

acceleration,a= qE/m = (1.6*10-19*1.5*103)/(9.11*10-31)

a= 2.634*1014 m/s2

v2 - u2 =2as

v2 - (-4.78*106)2 = 2* 2.634*1014 *0.39

v2 = 2.283*1014

v=1.510*107 m/s

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