HW 2 SPRING 18 x C Use The Worked Examp CI www.webassign.net/web/Student/Assignm
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HW 2 SPRING 18 x C Use The Worked Examp CI www.webassign.net/web/Student/Assignment-Responses/submit?dep. 181 63721 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 rield with magnitude 1.50 × 103 N/C pointing in the positive x-direction. (a) Assuming an initial speed of zero, nnd the speed of a proton at x = 0.0800 m with a potential energy of-2.40 × 10-17 . (Assume the potential energy at the point of release is zero.) 17 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 = 0.100 m, a change in potential energy of 2.88 x 10-17 9.1806 m/s EXERCISE HINTS: GETTING STARTED I IM STUck! The electron in part (b) travels from x 0.100 m (where it has half the initial speed you previously calculated) to x-0.240 m within the constant electric field. If there's a change in electric potential energy of -8.17 x 1017J as it goes from x- 0.100 m to x-0.240 m, find the electron's speed at x--0.240 m. (Note: Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.) m/s Need Help? Read It Submit Answer Save Progress macticeAnother Version] 1.05/1 points | Previous Answers SerCP11 16Q0.001 If an electron is released from rest in a uniform electric fleld, the electric potential energy of the charge-field 3. decreases remains the sameExplanation / Answer
EXERCISE:
electric potential energy decreases so it will gain in kinetic energy by same amount. (Energy consercvation)
delta(KE) = 8.17 x 10^-17
9.109 x 10^-31 ( v^2 - (9.18 x 10^6/2)^2) / 2 = 8.17 x 10^-17
v = 1.41 x 10^7 m/s .......Ans
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