Explanation:
It is given that,
Fundamental frequency, f = 220 Hz
(a) We know that at 0 degrees, the speed of sound in air is 331 m/s.
For open pipe,
l is the length of pipe
Also,
(b) Let f' is the fundamental frequency of the pipe at 30 degrees and v' is its speed.
v' = 348.71 m/s
So,
f' = 232.4 Hz
Hence, this is the required solution.
Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)
