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# old hw 5 - linares(jl36797 oldhomework 05 Turner(56705 This...

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linares (jl36797) – oldhomework 05 – Turner – (56705) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron begins at rest, and then is accel- erated by a uniform electric field of 800 N / C that extends over a distance of 7 cm. Find the speed of the electron after it leaves the region of uniform electric field. The elementary charge is 1 . 6 × 10 19 C and the mass of the electron is 9 . 11 × 10 31 kg. Correct answer: 4 . 43517 × 10 6 m / s. Explanation: Let : e = 1 . 6 × 10 19 C , m e = 9 . 11 × 10 31 kg , E = 800 N / C , and Δ x = 7 cm = 0 . 07 m . Because of the constant acceleration, v 2 = v 2 0 + 2 a Δ x . Since v 0 = 0 and a = F net m e = e E m e , v = radicalbigg 2 e E Δ x m e = radicalBigg 2 (1 . 6 × 10 19 C) (800 N / C) 9 . 11 × 10 31 kg × 0 . 07 m = 4 . 43517 × 10 6 m / s . 002 10.0 points A particle of mass 9 . 2 × 10 5 g and charge 28 mC moves in a region of space where the electric field is uniform and is 6 . 2 N / C in the x direction and zero in the y and z direction. If the initial velocity of the particle is given by v y = 5 . 7 × 10 5 m / s, v x = v z = 0, what is the speed of the particle at 0 . 6 s? Correct answer: 1 . 26756 × 10 6 m / s. Explanation: Let : m = 9 . 2 × 10 5 g = 9 . 2 × 10 8 kg , E x = 6 . 2 N / C , E y = E z = 0 , v y = 5 . 7 × 10 5 m / s , v x = v z = 0 , and t = 0 . 6 s . According to Newton’s second law and the definition of an electric field, vector F = mvectora = q vector E . Since the electric field has only an x compo- nent, the particle accelerates only in the x direction a x = q E x m . To determine the x component of the final velocity, v xf , use the kinematic relation v xf = v xi + a ( t f - t i ) = a t f . Since t i = 0 and v xi = 0 , v xf = q E x t f m v xf = (0 . 028 C) (6 . 2 N / C)(0 . 6 s) (9 . 2 × 10 8 kg) = 1 . 13217 × 10 6 m / s . No external force acts on the particle in the y direction so v yi = v yf = 5 . 7 × 10 5 m / s. Hence the final speed is given by v f = radicalBig v 2 yf + v 2 xf = bracketleftbigg ( 5 . 7 × 10 5 m / s ) 2 + ( 1 . 13217 × 10 6 m / s ) 2 bracketrightbigg 1 / 2 = 1 . 26756 × 10 6 m / s . Note: This is analogous to a particle in a gravitational field with the coordinates ro- tated clockwise by π 2 (90 ).

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linares (jl36797) – oldhomework 05 – Turner – (56705) 2 003 (part 1 of 2) 10.0 points A cone with base radius r and height h is
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