An electron of mass 9.11x10^-31 kg has an initial speed of 5.0x10^5 m/s. It travels in a straight line?

Question

An electron of mass 9.11x10^-31 kg has an initial speed of 5.0x10^5 m/s. It travels in a straight line and its speed increases to 8.00x10^5 m/s ina distance of 10cm. Assuming its acceleration is constant, determine the force exerted on the electron. And what is the ratio of this force to the weight of the electron?

Answer 1

constant acc. means that you can use the mean speed to calculate the time.
t=0.1/(8*10^5-5*10^5)/2= 6.666*10^-7s
acc= (8*10^5-5*10^5)/6.666*10^-7 =4.5*10^11m/s^2
Force=9.11*10^-31*4.5*10^11 =41*10^-20N
Ratio=41*10^-20/(9.11*10^-31) =4.5*10^11N/Kg

edited for mistakes. Hope it is right now.
edit2: OK I got the time wrong, and I don't want to edit again.Just take the distance/mean speed.

Answer 2

in case you artwork this out because of the fact the previous answerer did, you get a velocity quicker than gentle. So this question is improperly worded; it incredibly is going to say relax mass (m0) = 9.11 × 10-31 kg. the stunning formula for momentum, p, for speeds corresponding to the fee of light is p = gamma x mv the place gamma = a million /sqrt(a million-(v/c)^2) 3.sixty 4 × 10-22 = gamma x 9.11 × 10-31 x v this provides (gamma x v) = 3.sixty 4 × 10-22 / 9.11 × 10-31 = 4x10^8 m/s (close to sufficient). because c =3x10^8 m/s, this skill (gamma x v) = 4c/3. So the priority now's to locate v so (gamma x v) = 4c/3, i.e v / sqrt(a million-(v/c)^2) = 4c/3 v^2 / (a million-(v/c)^2) = 16c^2/9 a million - (v/c)^2 = 9v^2 / 16c^2 (c^2 - v^2)/c^2 = 9v^2/16c^2 c^2 - v^2= 9v^2/sixteen 25v^2/sixteen = c^2 v^2 = 16c^2/25 v = (4/5)c = 2,4 x 10^8 m/s

Answer 3

F = ma = (9.11)(10^-31)(8^2-5^2)(10^10)/(2*0.1) N
W = mg
F/W = a/g = (8^2-5^2)(10^10)/(2*0.1*9.8)