A proton moves at 4.20 times 10^5 m/s in the horizontal direction. It enters a u

Question

A proton moves at 4.20 times 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.20 times 10^3 N/C. Ignore any gravitational effects. Find the time interval required for the proton to travel 6.00 cm horizontally. 143 ns Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.) 9.21 mm Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally.

Explanation / Answer

here,

vx = 4.2 * 10^5 m/s

magnitude of electric feild , E = 9200 N/C

vertical accelration , a = q * E /m

a = 1.6 * 10^-19 * 9200 /(1.67 * 10^-27) m/s^2

a = 8.81 * 10^11 m/s^2

a)

time interval , t = 143 * 10^-9 s

b)

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

y = 0 + 0.5 * 8.81 * 10^11 * (143 * 10^-9)^2

y = 9 * 10^-3 m

c)

vertical velocity , vy = 0 + a * t

vy = 1.26 * 10^5 m/s

horizontal speed , vx = 4.2 * 10^5 m/s

the final velocity , v = ( 4.2 i + 1.26 j ) * 10^5 m/s

v = ( 420 i + 126 j ) km/s

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