A student standing 15m away kicks a ball off the ground towards the student kick

Question

A student standing 15m away kicks a ball off the ground towards the student kicks the ball with an initial velocity of 25m/s A student standing 15m away kicks a ball off the ground towards the student kicks the ball with an initial velocity of 25m/s A proton accelerates from rest in a uniform electric field of 680 N/C. At some later time, its speed is 1.08 x 106 m/s. (a) Find the magnitude of the acceleration of the proton 5.84e10 Calculate the force that the electric field exerts on the proton and use that force to calculate the acceleration of the proton.m/s (b) How long does it take the proton to reach this speed? 1.85e-5 Use the acceleration you calculated in part (a) of this probiem to find the time required for the proton to reach the specified speed. us (c) How far has it moved in that interval? 540200 Apply the kinematic equation for the distance traveled in a given time at uniform acceleration. m (d) What is its kinetic energy at the later time? 9026-22 Use the mass and final speed of the proton to calculate its final kinetic energy. Need Help? ResMaste

Explanation / Answer

(a) mass of the proton m = 1.673 x 10^-27 kg

charge of the proton, q = 1.602 x 10^-19 C

Force on the proton, F = q*E

So, acceleration a = F/m = q*E / m = (1.602 x 10^-19x680) / (1.673 x 10^-27) = 651.1 x 10^8 = 6.51 x 10^10 m/s^2

(b) Final speed, v = 1.08 x 10^6 m/s

use the expression -

v = u + a*t

=> 1.08 x 10^6 = 0 + 6.51 x 10^10 x t

=> t = (1.08 x 10^6) / (6.51 x 10^10) = 0.166 x 10^-4 s = 1.67 x 10^-5 s

(c) distance travelled, d = v^2 / (2*a) = (1.08 x 10^6)^2 / (2x6.51 x 10^10) = 8.3 m

(d) Kinetic energy KE = (1/2)*m*v^2 = 0.5*(1.673 x 10^-27)*(1.08 x 10^6)^2 = 0.976 x 10^-15 = 9.76 x 10^-16 J

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