In this example, suppose a proton (mp 1.67 x 10 27 kg) is released from rest at the positive plate. What is its speed just before it reaches the negative plate? Express your answer with the appropriate units. PA v=2.40.105 m S HAMO ?

In this example, suppose a proton (mp 1.67 x 10 27 kg) is released from rest at the positive plate. What is its speed just before it reaches the negative plate? Express your answer with the appropriate units. PA v=2.40.105 m S HAMO ?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

Part A - Practice Problem:
In this example, suppose a proton (mp = 1.67 × 10-27 kg) is released from rest at the positive
plate. What is its speed just before it reaches the negative plate?
Express your answer with the appropriate units.
HÅ
v=2.40 10³
●
m
S
?

100 V
The thin arrows represent
the uniform electric field.
dio
E
F=-E
1.0 cm

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so the distance is from -3 to 5 cm which is 8cm or .08m. I did parts a-e and need help for parts f-h. This is what I have so far: a: V1=1000V b: 3.2e^-16 J c. potential energy; no work needed to move particle as it leaves electric field d. 3.2e^-16 J e. m1<m2<m3<m4, v1>v2>v3>v4 f. distance =.08m -time takes particle 1 hit detector: .98e^-6 sec ---> .08m/.98e^-6 s= 8.16e^4 m/s -time takes particle 2 hit detector: 1.4e^-6 s --> .08m/1.4e^-6 s= 5.71e^4 m/s -time takes particle 3 hit detector: 1.7e^-6 s --> .08m/1.7^-6 s= 4.7053^4 m/s -time takes particle 4 hit detector: 1.96^-6 s --> .08m/1.96^-6 s= 4.082e^4 m/s I am not sure if i Made any mistakes so please let me know if i DO!