details please FULL SCREEN PRINTER VERSION BACK NEXT Chapter 22, Problem 054 In

Question

details please

FULL SCREEN PRINTER VERSION BACK NEXT Chapter 22, Problem 054 In the figure an electron is shot at an initial speed of vo-5.82 x 106 m/s, at angle eo-39.5?from an x axis. It moves through a uniform electric field (5.48 N/C) j. A screen for detecting electrons is positioned parallel to the y axis, at distance x 3.10 m. What is the y component of the electron's velocity (sign included) when the electron hits the screen? Detecting screen 0o Units Number the tolerance is +/-2% Click if you would like to Show Work for this question: udy Open Show Work

Explanation / Answer

mass of the electron, = 9.11 x 10^-31 kg

vo = 5.82 x 10^6 m/s

theta = 39.5 degree

E = 5.48 N/C j

x = 3.10 m

Acceleration of the electron,

a = F/m = eE / m = (-1.6 x 10^-19 x 5.48 j) / (9.11 x 10^-31)

a = -9.62 x 10^11 j

time taken for distance x = 3.10 m

t = x / ux = x / u*cos(theta)

t = 3.10 / (5.48 x 10^6 x cos(39.5))

t = 7.33 x 10^-7 sec

Horizontal component of the final velocity,

vx = u*cos(39.5) = 4.23 x 10^6 m/s

verticle component of the final velocity

vy = uy + ay*t

vy = (5.48 x 10^6 x sin(39.5) + ( -9.62 x 10^11 )* 7.33 x 10^-7 s

vy = 3.49 x 10^6 - 7.05 x 10^5

vy = 2.784 x 10^6

final velocity is

v = vx i + vy j

v = 4.23 x 10^6 m/s i + 2.784 x 10^6 m/s j

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