Particles q1 = -53.0 uc, q2 = +105 uc, and q3 = -88.0 uc are in a line. Particles qı and q2 are

separated by 0.50 m and particles q2 and q3 are

separated by 0.95 m. What is the net force on

particle qı?

Remember: Negative forces (-F) will point Left

Positive forces (+F) will point Right

-53.0 μC

-88.0 C

+105 με

+92

91

93

K 0.50 m

0.95 m

Enter

no

i wanna say it’s between b and d, leaning more towards b.

b) phosphorus pentoxide

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray