()226613222002149(4.24 10 m s)4.24 10 m s1.40 10 m ssinsin23.51 10 m s6.43 10 s.vvadtaθθ−×−×−×−−==×=×The negative root was used because we want the earliesttime for which y= d. The xcoordinate is ( )6920cos6.00 10 m s 6.43 10 s cos452.72 10 m.xvt××°=×θThis is less than Lso the electron hits the upper plate at x= 2.72 cm. 86. (a) The electric field is upward in the diagram and the charge is negative, so the force of the field on it is downward. The magnitude of the acceleration is a= eE/m, where Eis the magnitude of the field and mis the mass of the electron. Its numerical value is a=×××−−160 10911 10351 10193114....C2.0010NCkgms32chchWe put the origin of a coordinate system at the initial position of the electron. We take the xaxis to be horizontal and positive to the right; take the yaxis to be vertical and positive toward the top of the page. The kinematic equations are 201cos ,sin,andsin.2yxvty vtatvvat−=−First, we find the greatest
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.