With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 
N and 66 degrees
ELECTRIC FORCE (F)
F = 
Where K = 9 x 
N
/
The distance between 
and 
can be calculated by using Pythagoras theorem.
d = 
d = 46.7 cm = 0.467 m
For force 
, substitute all the parameters into the formula above
= (9 x x 3 x 1)/
= 2.7 x 
/0.218
= 1.24 x N
For force 
, substitute all the parameters into the formula above
= (9 x x 3 x 4)/
= 1.08 x /0.1089
= 9.92 x N
For force 
, substitute all the parameters into the formula above
= (9 x x 3 x 2)/
= 5.4 x /0.1089
= 4.96 x N
Summation of forces on Y component will be
= - Sin 45
= 9.92 x - 1.24 x Sin 45
= 9.04 x N
Summation of forces on X component will be
= - Cos 45
= 4.96 x - 1.24 x Sin 45
= 4.08 x N
Net Force = 
Net force = 
Net force = 9.9 x N
The direction will be
Tan ∅ = /
Tan ∅ = 9.04 x / 4.08 x
Tan ∅ = 2.216
∅ = 
(2.216)
∅ = 65.7 degrees
Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x N and 66 degrees approximately.
Learn more about electric Force here: brainly.com/question/4053816
Answer:
10.0 N West
Explanation:
Imagine East is positive and West is negative. Essentially now we have 2 forces pulling opposite each other on the same plane, so we can add them up.
F1=20N
F2=-30N
F1+F2=Net Force
20+(-30)=-10
Therefore the Net Force is 10N West.
Answer:
earth's shadow covering the moon,thats lunar eclipse
Answer:
0.687 m/s
Explanation:
Initial energy = final energy
1/2 mu² = mgh + 1/2 mv²
1/2 u² = gh + 1/2 v²
Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:
1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²
v = 0.687 m/s
