Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :
Wavelength for f = 45 Hz is,
Wavelength for f = 375 Hz is,
So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Answer:
The period would decrease by sqrt(2)
Explanation:
The restoring force is given by,
F = -kx
According to Newton's second law of motion,
ma = -kx
ma + kx = 0
The time period is given by,
T =
Where is the angular velocity and it is given by,
=
Now if the spring constant is doubled then,
Thus,
=
Thus, The period would decrease by sqrt(2).
Hence, option D is correct.
Answer:
t'=1.1897 μs
Explanation:
First we will calculate the velocity of micrometeorite relative to spaceship.
Formula:
where:
v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).
u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)
u' is the relative velocity of micrometeorite with respect to spaceship.
In order to find u' , we can rewrite the above expression as:
u'=0.9806c
Time for micrometeorite to pass spaceship can be calculated as:
(c = 3*10^8 m/s)
t'=1.1897 μs