Answer: T = 472.71 N
Explanation: The wire vibrates thus making sound waves in the tube.
The frequency of sound wave on the string equals frequency of sound wave in the tube.
L= Length of wire = 26cm = 0.26m
u=linear density of wire = 20g/m = 0.02kg/m
Length of open close tube = 86cm = 0.86m
Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as
L = 3λ/4
0.86 = 3λ/4
3λ = 4 * 0.86
3λ = 3.44
λ = 3.44/3 = 1.15m.
Speed of sound in the tube = 340 m/s
Hence to get frequency of sound, we use the formulae below.
v = fλ
340 = f * 1.15
f = 340/ 1.15
f = 295.65Hz.
f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.
The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as
L = λ/2
0.26 = λ/2
λ = 0.52m
The speed of sound in string is given as v = fλ
Where λ = 0.52m f = 295.65 Hz
v = 295.65 * 0.52
v = 153.738 m/s.
The velocity of sound in the string is related to tension, linear density and tension is given below as
v = √(T/u)
153.738 = √T/ 0.02
By squaring both sides
153.738² = T / 0.02
T = 153.738² * 0.02
T = 23,635.372 * 0.02
T= 472.71 N
Answer:
did you ever get the answer
Answer:
<em>The resultant velocity has a magnitude of 38.95 m/s</em>
Explanation:
<u>Vector Addition</u>
Given two vectors defined as:


The sum of the vectors is:

The magnitude of a vector can be calculated by

Where x and y are the rectangular components of the vector.
We have a plane flying due west at 34 m/s. Its velocity vector is:

The wind blows at 19 m/s south, thus:

The sum of both velocities gives the resultant velocity:

The magnitude of this velocity is:


d = 38.95 m/s
The resultant velocity has a magnitude of 38.95 m/s
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by

where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is

And the frequency of the motion instead is given by:

(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:

where
F is the force applied initially to the spring, so it is equal to the weight of the block:

k = 160 N/m is the spring constant
Solving for A, we find
