An electron with speed v0 = 5.00 x 10^6 m/s is traveling parallel to an electric field of magnitude e = 9900 n/c . 1)how far will the electron travel before it stops?
express your answer to three significant figures and include the appropriate units.
2)how much time will elapse before it returns to its starting point?
express your answer to three significant figures and include the appropriate units

Explanation:

in the electric field electron will face a force which will create an acceleration

( here - ve ) as follows

Force on electron

= charge on electron x electric field  = Q X E

acceleration=  Force / mass = Q E / m

mass of electron = 9.1 x 10⁻³¹

acceleration a =

= 17.4 x 10¹⁴ ms⁻² .

Now initial velocity  u = 5 x 10⁶ m/s

Final velocity v = 0

acceleration a = 17.4 x 10¹⁴ ms⁻²

distance of travel = s

v² = u² - 2as

0 = (5 x 10⁶)² - 2 x 17.4 x 10¹⁴ s

s = 7.18 mm

2 ) v = u - at

0 = 5 x 10⁶ - 17.4 x 10¹⁴ t

t = .287 x 10⁻⁸ s

Total time elapsed  = 2 x .287 x 10⁻⁸

= .57 x 10⁻⁸ s .

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2