A 160 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm and released. A student with a stopwatch finds
1 answer:
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In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 ×
)(1.2 ×
)
PE = 5.0 x joules
None of the options shown is correct.
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.