Answer:
The acceleration would be 3.455.
Answer: 0°
Explanation:
Step 1: Squaring the given equation and simplifying it
Let θ be the angle between a and b.
Given: a+b=c
Squaring on both sides:
... (a+b) . (a+b) = c.c
> |a|² + |b|² + 2(a.b) = |c|²
> |a|² + |b|² + 2|a| |b| cos 0 = |c|²
a.b = |a| |b| cos 0]
We are also given;
|a+|b| = |c|
Squaring above equation
> |a|² + |b|² + 2|a| |b| = |c|²
Step 2: Comparing the equations:
Comparing eq( insert: small n)(1) and (2)
We get, cos 0 = 1
> 0 = 0°
Final answer: 0°
[Reminders: every letters in here has an arrow above on it]
Answer:
Electric field E = kQ/r^2
Distance between charges = 6.30 - (-4.40) = 10.70m
Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.
Field from q1 at P = k(-9.50x^10^-6) / d^2
Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2
These fields are in opposite directions and are equal magnitudes if the resultant field = 0
k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2
9.50 / d^2 =8.40 / (10.70-d)^2
d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131
d/(10.70-d) = sqrt(1.1331) = 1.063
d = 1.063 ((10.70-d)
= 10.63 - 1.063d
2.063d = 10.63
d = 5.15m
The y coordinate where field is zero is 6.30 - 5.15 = 1.15m
Explanation:
I believe the answer is C
