Thank you in advance A proton moves at 5.20 x 10 m/s in the horizontal direction

Question

Thank you in advance

A proton moves at 5.20 x 10 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.00 x 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 5.50 cm horizontally. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully, ns (b) Find its vertical displacement during the time interval in which it travels 5.50 cm horizontally. (Indicate direction with the sign of your answer.) 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 It Master It Subrmit Answer Save ProgressPractice Another Version

Explanation / Answer

here,

horizontal speed , vx = 5.2 * 10^5 m/s

magnitude of electric feild , E = 9 * 10^3 N/C

horizontal distance , x = 5.5 cm = 0.055 m

a)

the time interval required to travel 5.5 cm horizontally , t = x/vx

t = 0.055 /( 5.2 * 10^5 ) s

t = 1.06 * 10^-7 s

b)

accelration , a = e * E/mp

a = 1.6 * 10^-19 * 9 * 10^3 /(1.67 * 10^-27 )

a = 8.62 * 10^11 m/s^2

the vertical displacement , y = 0 + 0.5 * a * t^2

y = 0.5 * 8.62 * 10^11 * ( 1.06 * 10^-7)^2 m

y = 4.84 * 10^-3 m

c)

final vertical speed , vy = 0 + a * t

vy = 0 + 8.62 * 10^11 * 1.06 * 10^-7 m/s

vy = 9.1 * 10^4 m/s

so, v = vx i + vy j

v = 5.2 * 10^5 i m/s + 9.1 * 10^4 j m/s

v = 520 i m/s + 91 j km/s

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