If the mass of a material is 76 grams and the volume of the material is 14 cm3, what would the density of the material be?
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
.
Complete Question
A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?
Answer:
The values is
Explanation:
From the question we are told that
The distance it travels eastward is
The distance it travel towards the southeast is
The distance it travel towards the south is
Let x-axis be east
y-axis south
z-axis into the ground
The angle made between east and south is
The displacement toward x-axis is
The displacement toward the y-axis is
Now the overall displacement of the rat is mathematically evaluated as
The initial height of the first body is given by:
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.