We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
The sound gets louder as it gets closer and when it passes is gets softer
His law exaplins/shows that the average distance of a planet from the Sun cubed is directly proportional to the orbital period squared.
Answer:
The frequencies are 
Explanation:
From the question we are told that
The length of the ear canal is 
The speed of sound is assumed to be 
Now taking look at a typical ear canal we see that we assume it is a closed pipe
Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as
substituting values
Also the the second harmonic for the pipe (ear canal) is mathematically represented as
substituting values
Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies
Answer:
0.657 seconds
Explanation:
speed of wave= wavelength / time period
so
time period= wavelength / speed
= 4.6/7
=0.657 sec
