answer question 14,15,&16 please 0 101 x 10g o None of the above. QUESTION 14 An

Question

answer question 14,15,&16 please

0 101 x 10g o None of the above. QUESTION 14 An electron is placed in a unftorm electric fed, E-200 +-100 Whatis the x component of the electron's acceleration? component ofthe electro 'sacos eration O -393 x 10 O 3.51x10' O 393 x 10u D Nore of the above. QUESTION 15 Consider the electron from the previous question. What direction is the electron accelerated? e 153.43 with respect to the +x axis. O 63.13 with respect to the +x axis. 243.43' with respect to the +x axis. O 333.43 with respect to the +x axis. o None of the above. QUESTION 16 Consider the electron from the previous question. How fast will the electron be traveling after i0.00js, if it was released from rest? 958 x 10 O 126 x 10 m 0 192 x 10 m 3x 10 None of the above. QUESTION 17 Click Save and Submit to save and submit. Click Save All Answers to save all ansuwers Type here to search

Explanation / Answer

14)

ax = q*Ex/m

q = -1.602*10^-19 C

m = 9.11*10^-31 Kg

Ex = 2.00 N/C

so,

ax = q*Ex/m

= (-1.602*10^-19)*(2.00)/(9.11*10^-31)

= -3.517*10^11 m/s^2

Answer: option 1

15)

similarly:

ay= q*Ey/m

= (-1.602*10^-19)*(-1.00))/(9.11*10^-31)

= 1.759*10^11 m/s^2

magnitude of a = sqrt ((-3.517*10^11)^2 + (1.759*10^11)^2)

magnitude of a = 3.93*10^11 m/s^2

Since x component is negative and y component is positive

It lies in the 2nd cordinate

angle = 90 + atan(|x/y|)

angle = 90 + atan(|-3.517*10^11/1.759*10^11|)

angle = 153.43 degree

Answer: option 1

16)

vf = vi +a*t

= 0 + (3.93*10^11)*10.00*10^-6

= 3.93*10^6 m/s

Answer: option 4

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