HW 1 - segovia (cs39966) Homework 1 Spurlock (14441) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: segovia (cs39966) Homework 1 Spurlock (14441) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Calculate the net charge on a substance con- sisting of a combination of 8 . 6 10 13 protons and 6 . 8 10 13 electrons. Correct answer: 2 . 88 10 6 C. Explanation: Let : N p = 8 . 6 10 13 charges , q p = 1 . 60 10 19 C , N e = 6 . 8 10 13 charges , and q e = 1 . 60 10 19 C . Q net = N p q p + N e q e Q net = (8 . 6 10 13 )(1 . 6 10 19 C) + (6 . 8 10 13 )( 1 . 6 10 19 C) = 2 . 88 10 6 C 002 10.0 points A particle of mass 41 g and charge 64 C is released from rest when it is 78 cm from a second particle of charge 17 C. Determine the magnitude of the initial ac- celeration of the 41 g particle. Correct answer: 392 . 008 m / s 2 . Explanation: Let : m = 41 g , q = 64 C = 6 . 4 10 5 C , d = 78 cm = 0 . 78 m , Q = 17 C = 1 . 7 10 5 C , and k e = 8 . 9875 10 9 . The force exerted on the particle is F = k e | q 1 || q 2 | r 2 = m a bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl m d 2 = k e vextendsingle vextendsingle 6 . 4 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 1 . 7 10 5 C vextendsingle vextendsingle (0 . 041 kg) (0 . 78 m 2 ) = 392 . 008 m / s 2 . 003 10.0 points Three charges are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). 5 nC 4 nC 9 nC y ( m ) 0 1 2 3 4 5 6 7 8 9 10 x ( m ) 1 2 3 4 5 6 7 8 9 10 What is the magnitude of the resulting force on the 5 nC charge at the origin? Correct answer: 12 . 8898 nN. Explanation: Let : q o = 5 10 9 C , ( x o , y o ) = (0 m , 0 m) , q a = 4 10 9 C , ( x a , y a ) = (4 m , 0 m) , q b = 9 10 9 C , and ( x b , y b ) = (0 m , 8 m) . Coulombs Law for q o and q a is bardbl vector F oa bardbl = k e q o q a radicalbig ( x a x o ) 2 + ( y a y o ) 2 = k e q o q a radicalbig x 2 oa + y 2 oa = 8 . 98755 10 9 N C 2 / m 2 segovia (cs39966) Homework 1 Spurlock (14441) 2 ( 5 10 9 C) ( 4 10 9 C) radicalbig (4 m) 2 + (0 m) 2 = 1 . 12344 10 8 N . F oa x = k e q o q a r 2 oa cos oa = k e q o q a r 2 oa x oa r oa = (8 . 98755 10 9 N C 2 / m 2 ) ( 5 10 9 C) ( 4 10 9 C) (4 m) 2 4 m 4 m = 1 . 12344 10 8 N . F oa y = k e q o q a r 2 oa sin oa = k e q o q oa r 2 oa y oa r oa = (8 . 98755 10 9 N C 2 / m 2 ) ( 5 10 9 C) ( 4 10 9 C) (4 m) 2 0 m 4 m = 0 N ....
View Full Document

Page1 / 18

HW 1 - segovia (cs39966) Homework 1 Spurlock (14441) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online