A proton moves at 5.00 × 105 m/s in the horizontal direction. It enters a unifor

Question

A proton moves at 5.00 × 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8, 20 103 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.) inin (c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally km/s

Explanation / Answer

a) in horizontal direction there is no net force. so, proton moves with constant speed.

t = d/v = 0.045/5.0*10^5 = 90.0*10^-9 s = 90.0 ns

b) F = q*E

m*a = q*E

a = q*E/m = 1.6*10^-19*8.2*10^3/1.67*10^-27 = 7.85*10^11 m/s^2

y = 0.5*a*t^2

= 0.5*7.85*10^11*(90.0*10^-9)^2

= 3.17*10^-3 m

= 3.17 mm

c) vx = 500 km/s

vy = a*t = 7.85*10^11*90.0*10^-9 = 70.650 km/s


V = (500 i + 70.650 j) km/s

I hope help you.....

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