A proton moves at 4.30 times 105 m/s in the horizontal direction. It enters a un

Question

A proton moves at 4.30 times 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 times 103 N/C. (Ignore any gravitational effects in your calculations.) Find the time interval required for the proton to travel 6.25 cm horizontally. Deltat = s Find its vertical displacement during the time interval in which it travels 6.25 cm horizontally. h = m Find the vertical component of its velocity after it has traveled 6.25 cm horizontally. vy = m/s Find the horizontal component of its velocity after it has traveled 6.25 cm horizontally. vx = m/s

Explanation / Answer

a) since field is vertical vx doesnt change vx=d/t t=d/vx=6.25E-2/4.3E5=1.45E-7 s b) y = y0 + v0y t + 1/2 a t^2 a= qE/m = 1.6E-19 * 8.4E3 / 1.67E-27=8.05E11 y=0+0t + 1/2 * 8.05E11 * (1.45E-7)^2=8.46 E-3 m c) v = v0+ at v= 0 + 8.05E11 * 1.45E-7=1.167E5 d) horizontal velocity doesnt change 4.3E5

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