x (SIU Homepage PHYS-255B x y 205B Homework No. 01 x C Search Textbook Solu x M
ID: 1524087 • Letter: X
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x (SIU Homepage PHYS-255B x y 205B Homework No. 01 x C Search Textbook Solu x M Mathway IMath Problen x SalukiNet C www.webassign.net/web/Student/Assignment-Responses/last?dep 15617266 Need Help L Read It 10. O 14 points I Previous Answers SerpSE9 23.P057.MI My Notes Ask Your Teacher A proton moves at 5.20 x 105 m/s In the horizontal direction It enters a uniform vertical electric field with a magnitude of 8.20 x 103 N/C. Ignore any gravitational effects. a) Find the time interval required for the proton 5.50 cm horizontally. o travel 105.7 (b) Find its vertical displacement during the time Interval n which it travels 5.50 cm horizontally. (Indicate direction with the sign of your answer.) 80441 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. mm (c) Find the horizontal and vertical components of its velocity after it has traveled 5.50 cm horizontally. j) km/s Need Help? Read t Master It Submit Assignment Save Assignment Progress Home My Assignments Extension Request WobAssigne 4.0 1997-2017 Advanced Instructional Systems, Inc. All rights reserved.Explanation / Answer
(b). In vertical direction :
Initial velocity u = 0
Time t = 105.7 ns = 105.7 x10 -9 s
Electric field E = 8.2 x10 3 N/C
Accleration a = Eq / m
= (8.2 x10 3)(1.6 x10 -19 ) /(1.67x10 -27 )
= 7.856x10 11 m/s 2
Vertical displacment S = ?
From the relation S = ut +(1/2) at 2
= 0 +[0.5x7.856x10 11x (105.7x10 -9 ) 2]
= 4.388x10 -3 m
(c). Final Vertical velocity v = u + at
= 7.856x10 11 x105.7x10 -9
= 83.037x10 3 m/s
So, velocity = (5.2x10 5 ) i + (83.037x10 3) j
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