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romanna
3 months ago
12

Driving down a quiet street on a summer day with the car windows down, you see a convertible approaching from the opposite direc

tion, with the driver playing her car stereo loudly. Having a well-trained musician’s ear, you notice that the song you hear is in a key one semitone (the interval between two adjacent keys on a piano) above the key in which the song is typically heard. You also know that to shift any note by one semitone, the frequencies must be related by a factor of 1.06. Assume both drivers are traveling at the same speed, what is their speed?
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
7 0
The ratio of the perceived frequency to the actual frequency is 1.06. Denote their speeds as v. Based on the Doppler effect formula applicable to a sound source and listener moving toward each other: n = n₀ (V + v) / (V - v) where V represents the speed of sound at 340 m/s. Thus, n / n₀ = (V + v) / (V - v), which leads to 1.08 = (360.4 - 1.06 v)/(340 + v); solving yields v = 9.9 m/s.
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The wavelength of light is 5000 angstrom. Express it in nm and m.
serg [3582]

Answer:

1 angstrom equals 0.1 nm.

To convert 5000 angstroms: 5000 angstrom = 5000 / 1 × 0.1 nm.

= 500 nm

1 \:  angstrom = 1 \times  {10}^{ - 10} m

To express 5000 angstroms in meters: 5000 angstrom = 5000 × 1 × 10^-10.

= 5 × 10^-7 m

Hope this explanation is useful for you.

7 0
4 months ago
An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
inna [3103]

When air is forced into the open pipe,

L = \frac{nλ}{2}

where n represents any whole number such as 1,2,3,4, etc., and λ denotes the wavelength of the oscillation

This implies λ=\frac{2L} {n}

It is important to note that n=1 corresponds to the fundamental frequency, n=2 corresponds to the first harmonic, and so forth.

Thus, the third harmonic will be for n=4

With L=6m and n=4, solving for λ yields:

λ=\frac{(2)*(6)}{4} =3m

The connection between frequency (f), sound speed (c), and wavelength (λ) is given by:

c=f.λ or f= \frac{c}{λ}

Therefore, f=\frac{344}{3}

≈115 Hz

8 0
3 months ago
Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
Ostrovityanka [3204]

Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

\theta _2=60^{\circ}, \theta _1, and \theta _2 are complementary angles which means their ranges are identical.

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

7 0
3 months ago
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