Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F = 
q = 
we calculate
q = 
q = 
Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer:
about 14.7°
Explanation:
The formula for the angle of the first minimum is ...
sin(θ) = λ/a
where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.
The wavelength of the sound is the speed of sound divided by the frequency:
λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m
Then the angle of interest is ...
θ = arcsin(0.261538/1.03) ≈ 14.7°
At an angle of about 14.7°, someone outside the room will hear no sound.
Answer:
Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for 
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by on q. Then we can know the magnitude of the force exerted by 
about q, finally this will allow us to know the magnitude of
exerts a force on q in +y direction, and exerts a force on q in -y direction.
The net force on q is:
Rewriting for :
