A proton moves at 4.20 × 105 m/s in the horizontal direction. It enters a unifor

Question

A proton moves at 4.20 × 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 7.80% 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 5.50 cm horizontally. 131 ns (b) Find its vertical displacement during the time interval in which it travels 5.50 cm horizontally. (Indicate direction with the sign of your answer.) 6.41 (c) Find the horizontal and vertical components of its velocity after it has traveled 5.50 cm horizontally. I 10.51 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. j) km/s

Explanation / Answer

vx = 4.20 x 10^5 m/s

ay = q E / m = (1.6 x 10^-19)(7.80 x 10^3)/(1.67 x 10^-27)

ay = 7.47 x 10^11 m/s^2

vy^2 - 0^2 = 2 (7.47 x 10^11) (0.0550)

vy = 2.87 x 10^5 m/s

v = (4.20 x 10^5) i + (2.87 x 10^5) j m/s

Ans: v = 420 i + 287 j km/s

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F = k x and F = ke q1 q2 / r^2

k (0.0384) = ( 9 x 10^9) (0.760 x 10^-6)(0.642 x 10^-6)/0.052^2

k = 42.3 N/m ....Ans

12. (D) KE = m v^2 / 2

= (1.67 x 10^-27) (1.30 x 10^6)^2 / 2

= 1.41 x 10^-15 J

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