Answer:
0.0567 N
Explanation:
q1 = 3 micro coulomb
q2 = - 2 micro coulomb
OB = 50 cm
AB = 60 cm
By using Pythagoras theorem in triangle OAB
OA = 78.1 cm
By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is
F = 0.0885 N
The horizontal component of force is
= F CosФ =
= 0.0885 x 50 / 78.1 = 0.0567 N
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be = ..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength = ..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance = ..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm