1) An electron is ejected horizontally at a speed of 1.5 106 m/s from the electr

Question

1) An electron is ejected horizontally at a speed of 1.5 106 m/s from the electron gun of a computer monitor. If the viewing screen is 35 cm away from the end of the gun, how far will the electron travel in the vertical direction before hitting the screen?


2) A stone thrown off a bridge 20 m above a river has an initial velocity of 12 m/s at an angle of 450 above the horizontal.
a. What is the range of the stone? ___________
b. At what velocity does the stone strike the water?
Magnitude _______
Direction ________ below the horizontal

Explanation / Answer

Given 1)Horizontal velocity, vxi = 1.5 *10^6 m/s Distance ,x =0.35 m First we will calculate the time taken x = vxi t 0.35 m = (1.5 *10^6 m/s) t t =0.23 *10^-6 s The vertical distance travelled by the electron is y =1/2 gt^2 = 0.5 *9.8 m/s^2 *(0.23 *10^-6 s)^2 = 0.25 *10^-12 m ----------------------------------------------------------------------- 2)Height of the bridge, h =20 m Initial velocity, vi =12 m/s Angle of projection,? =45^0 a)Range = vi^2 sin2?/g          = (12 m/s)^2 sin90/9.8 m/s^2 Range = 14.69 m ------------------------------------------------------------ b)The velocity of stone is     vxi = vxf = vi cos45                  = 12 *0.707 =8.48 m/s     vyf^2-vyi^2 = 2as      vyf^2 -(vyi sin45)^2 = 2 *9.8 m/s^2 *20 m    vyf =21.5 m/s The magnitude of velocity is v = v(8.48)^2+(21.5)^2 =23.1 m/s Direction cos? = 8.48/23.11 ? = 68.47^0 below the horizontal

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