Sunday

March 29, 2015

March 29, 2015

Posted by **carlton** on Monday, October 17, 2011 at 10:47am.

(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?

- physics -
**Damon**, Monday, October 17, 2011 at 11:07amm = 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

- physics -
**drwls**, Monday, October 17, 2011 at 11:10am(a) Force * distance = kinetic energy increase

Solve for the force.

(b) Divide first answer by m*g, where m is the electron mass.

**Answer this Question**

**Related Questions**

physics - An electron of mass 9.11 10-31 kg has an initial speed of 2.85 105 m/s...

physics - An electron has an initial speed of 2.95 × 105 m/s.If it undergoes an ...

Physics - A 747 jetliner lands and begins to slow to a stop as it moves along ...

physics - An electron ( mass = 9.11×10−31 kg) leaves one end of a TV ...

Maths - Suppose that a boat moves in a straight line with speed v through water ...

physics - A 747 jetliner lands and begins to slow to a stop as it moves along ...

Physics - Two bicycle tires are set rolling with the same initial speed of 3.00m...

PHYSICS LEC - please help...i need this immediately.. rolling friction. two ...

physics - A proton (mass ) is being accelerated along a straight line at in a ...

Physics - A proton (mass m = 1.67 10-27 kg) is being accelerated along a ...