The periodic table lists all the elements according to the number of __________ they have! You can’t change __________ without c
1 answer:
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Answer:
Explanation:
(A)
The string has set of normal modes and the string is oscillating in one of its modes.
The resonant frequencies of a physical object depend on its material, structure and boundary conditions.
The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.
Given below are the incorrect options about the wave in the string.
• The wave is travelling in the +x direction
• The wave is travelling in the -x direction
• The wave will satisfy the given boundary conditions for any arbitrary wavelength
• The wave does not satisfy the boundary conditions
Here, the string of length L held fixed at both ends, located at x=0 and x=L
The key constraint with normal modes is that there are two spatial boundary conditions,
and
.The spring is fixed at its two ends.
The correct options about the wave in the string is
• The wavelength can have only certain specific values if the boundary conditions are to be satisfied.
(B)
The key factors producing the normal mode is that there are two spatial boundary conditions, and
, that are satisfied only for particular value of
.
Given below are the incorrect options about the wave in the string.
• must be chosen so that the wave fits exactly o the string.
• Any one of or
or
can be chosen to make the solution a normal mode.
Hence, the correct option is that the system can resonate at only certain resonance frequencies and the wavelength
must be such that
(C)
Expression for the wavelength of the various normal modes for a string is,
(1)
When , this is the longest wavelength mode.
Substitute 1 for n in equation (1).
When , this is the second longest wavelength mode.
Substitute 2 for n in equation (1).
When , this is the third longest wavelength mode.
Substitute 3 for n in equation (1).
Therefore, the three longest wavelengths are ,
and
.
(D)
Expression for the frequency of the various normal modes for a string is,
For the case of frequency of the normal mode the above equation becomes.
Here, is the frequency of the
normal mode, v is wave speed, and
is the wavelength of
normal mode.
Therefore, the frequency of normal mode is
.
Answer:
Acceleration of Sea Lion is 4.41 g
This is 49% of maximum jet acceleration given as a = 9g
Explanation:
As we know that the radius of the circular loop is given as
R = 0.37 m
The speed of the fish is given as
Now the centripetal acceleration of the sea lion is given as
as we know that
so we have
Now Percentage of this acceleration wrt maximum jet acceleration is given as
%