Asked by carlton
An electron of mass 9.11 10-31 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to 8.00 105 m/s in a distance of 10.0 cm.
(a) Assuming its acceleration is constant, determine the force exerted on the electron.
N (in the direction of motion)
(b) What is the ratio of this force to the weight of the electron, which we neglected?
(a) Assuming its acceleration is constant, determine the force exerted on the electron.
N (in the direction of motion)
(b) What is the ratio of this force to the weight of the electron, which we neglected?
Answers
Answered by
Damon
m = 9.11 * 10^-31
vi = 3 * 10^5
vf = 8 * 10^5
vf = vi + a t
d = Vi t + .5 a t^2
10^-1 = 3*10^5 t + .5 a t^2
or
1 = 3*10^6 t + 5 a t^2
but
t = (8-3)10^5 /a
so
1 = 3*10^6 (5*10^5/a ) +5 (25*10^10/a)
a = 15*10^11 + 12.5*10^11
a = 27.5 * 10^11 kg
F = ma = 9*10^-31*27.5*10^11
= 247.5 * 10^-20
= 2.475 * 10-22 N
weight of electron = 9.81*9.11*10^-31
so do ratio
vi = 3 * 10^5
vf = 8 * 10^5
vf = vi + a t
d = Vi t + .5 a t^2
10^-1 = 3*10^5 t + .5 a t^2
or
1 = 3*10^6 t + 5 a t^2
but
t = (8-3)10^5 /a
so
1 = 3*10^6 (5*10^5/a ) +5 (25*10^10/a)
a = 15*10^11 + 12.5*10^11
a = 27.5 * 10^11 kg
F = ma = 9*10^-31*27.5*10^11
= 247.5 * 10^-20
= 2.475 * 10-22 N
weight of electron = 9.81*9.11*10^-31
so do ratio
Answered by
drwls
(a) Force * distance = kinetic energy increase
Solve for the force.
(b) Divide first answer by m*g, where m is the electron mass.
Solve for the force.
(b) Divide first answer by m*g, where m is the electron mass.
Answered by
seto
thank you
Answered by
samic
thank u
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