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scZoUnD [109]
3 years ago
9

A speaker at the front of a room and an identical speaker at the rear of the room are being driven at 456 Hz by the same sound s

ource. A student walks at a uniform rate of 1.12 m/s away from one speaker and toward the other. How many beats does the student hear per second? (Take the speed of sound to be 343 m/s.) Hint: The Doppler effect causes both frequencies to be shifted. The difference between those two frequencies is what causes the beats.
Physics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

2.97795 Hz

Explanation:

v = Speed of sound in air = 343 m/s

v_r = Relative speed between the speakers and the student = 1.12 m/s

f' = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.12}{343}\\\Rightarrow f_1=454.51102\ Hz

Frequency of sound heard as the student moves closer to the other speaker

f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.12}{343}\\\Rightarrow f_2=457.48897\ Hz

The difference in the frequencies is

f_2-f_1=457.48897-454.51102=2.97795\ Hz

The student hears 2.97795 Hz

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A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
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Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

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a= 130.3m/s²

3 0
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Section 2.2
Feliz [49]

Answer:

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Explanation:

We are told the object's acceleration is equal to 24 m/s²

Now we know that acceleration can also be defined as the rate of change of speed with time. Also speed has a unit known as m/s.

Thus, we can rephrase the acceleration in this question to mean;

Speed changes at the rate of 24 m/s for every second with time.

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