A proton moves at 5.00 times 10^5 m/s in the horizontal direction. It enters a u

Question

A proton moves at 5.00 times 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.80 times 10^3 N/C. Ignore any gravitational effects. Find the time interval required for the proton to travel 4.50 cm horizontally. _____________ ns Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally. (Indicate direction with the sign of your answer.) ________________ mm Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally. v vector=

Explanation / Answer

horizontal velocity will remain constant

speed = distance / time

5 * 10^5 = 0.045 / time

time = 9 * 10^-8 sec or 90 ns

acceleration = charge * electric field / mass

acceleration = 1.6 * 10^-19 * 7.8 * 10^3 / (1.672 * 10^-27)

acceleration = 746411483254 m/s^2

s = ut + 0.5 * at^2

s = 0 + 0.5 * 746411483254 * (90 * 10^-9)^2

vertical displacement = 0.00302 m

vertical dispacement = 3.02 mm

v = u + at

vertical velocity = 0 + 746411483254 * (90 * 10^-9)

vertical velocity = 67177.03 m/s

v = (5 * 10^2i + 67.17703j) km/s

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