A proton moves at 3.60 x 105 m/s in the horizontal direction. It enters a unifor

Question

A proton moves at 3.60 x 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.00 x 10 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.50 cm horizontally. ns (b) Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally, (Indicate direction with the sign of your answer.) (c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 an horizontally. j) km/s

Explanation / Answer

here,

a) horizontal displacement s=4.50 cm =0.045 m

when the proton horizontally travelled in vertical electic feild, there is no effect in motion of proton

the horizontal velocity of the proton constant through out the motion

therefore velocity v =displacement / time

time t =displacement /velocity = 0.045/ 3.6*10^5

t = 1.25 * 10^-7 s

b)

vertical accelration , a = e * E /mp

a = 1.6 * 10^-19 * 9000 /( 1.67 * 10^-27) = 8.6175 * 10^11 m/s^2 j

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

y = 0.5 * 8.6175 * 10^11 * ( 1.25 * 10^-7)^2

y = 6.732 * 10^-3 m

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