Answer:
a) L = 1,923 m
, b) L = 0.012 m
Explanation:
This is a resonance problem, if the tube is open at both ends it has a
a maximum in each one, the waves must fulfill the relationship
λ = 2L / n
Where n is an integer and L the length of the tube
Let's use the ratio of the speed of sound
v = λ f
v = 2L f / n
L = v/2f n
The fundamental frequency corresponds to n = 1
L = v / 2f
a) for the smallest frequency
f = 89.2 Hz
L = 343 / (2 89.2)
L = 1,923 m
b) for the highest frequency
f = 13.9 kHz = 13.9 10³ Hz
L = 343 / (2 13.9 10³)
L = 12.338 10⁻³ m
L = 0.012 m
Answer:
The speed of the wave remains the same
Explanation:
Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.
We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.
<u>So, If you increase the frequency of oscillations, the speed of the wave remains the same.</u>
Answer: 
Explanation:
The Compton Shift 
in wavelength when the photons are scattered is given by the following equation:
(1)
Where:
is a constant whose value is given by 
, being 
the Planck constant, 
the mass of the electron and 
the speed of light in vacuum.
the angle between incident phhoton and the scatered photon.
We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 
:
(2)
(3)
Now, let's find the angle that will produce a fourth of this maximum value found in (3):
(4)
(5)
If we want 
, 
must be equal to 1:
(6)
Finding 
:
Finally:
This is the scattering angle that will produce 
Answer:31.62 m/s
Explanation:
Given
mass of body 
Pull on chain is 
Pull get smaller at the rate of 
Net Upward Force 
net acceleration 
but g is acting downward
using 
here initial velocity is zero
