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:
a. 16 s b. -1.866 kJ
Explanation:
a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.
We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².
Since the turntable stops at ω = 0, the time it takes to stop is gotten from
ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.
So it takes the turntable 16 s to stop.
b. The workdone by the turntable to stop W equals its rotational kinetic energy change.
So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ
Answer:
height of a cliff is 45.86 m
Explanation:
u=8 m/s
t=2.35 s
the height of the cliff is
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