F = m*a, mass times acceleration.
F = 15*10 = 150 N
Answer:
71.4583 Hz
67.9064 N
Explanation:
L = Length of tube = 1.2 m
l = Length of wire = 0.35 m
m = Mass of wire = 9.5 g
v = Speed of sound in air = 343 m/s
The fundamental frequency of the tube (closed at one end) is given by

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz
The linear density of the wire is

The fundamental frequency of the wire is given by

The tension in the wire is 67.9064 N
Answer:
is reflected back into the region of higher index
Explanation:
Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.
According to Snell's law, refraction of ligth is described by the equation

where
n1 is the refractive index of the first medium
n2 is the refractive index of the second medium
is the angle of incidence (in the first medium)
is the angle of refraction (in the second medium)
Let's now consider a situation in which

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

Where
is a number greater than 1. This means that above a certain value of the angle of incidence
, the term on the right can become greater than 1. So this would mean

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

Q stands for heat energy (Joules) hope this helps!