A band director instructs students to play a pitch louder. How will the sound wave change when the band plays the same pitch lou
der?
The speed of the sound wave will increase.
The frequency of the sound wave will increase.
The wavelength of the sound wave will increase.
The amplitude of the sound wave will increase.
The speed of the sound wave will increase.
The frequency of the sound wave will increase.
The wavelength of the sound wave will increase.
The amplitude of the sound wave will increase.
2 answers:
Explanation :
Loudness is defined as how our ear perceives the sound. It is basically related to the intensity of a sound wave.
When the same pitch is played louder, its amplitude changes. Amplitude is the maximum displacement covered by a wave.
This is the only difference between the loudness to that of shrillness. The amplitude of the loud sound is more as compared to that of shrill sound.
So, the correct option is (d).
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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
The angular magnification is
Explanation:
From the question we are told
The focal length is
The near point is
The angular magnification is mathematically represented as
Substituting values