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Mandarinka [93]

2 years ago

7

A loud factory machine produces sound having a displacement amplitude in air of 1.00 μmμm, but the frequency of this sound can b

e adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 PaPa. Under the conditions of this factory, the bulk modulus of air is 1.30×105 PaPa . The speed of sound in air is 344 m/s
What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?

1 answer:

lozanna [386]2 years ago

3 0

Answer:

f =3.4*10^3 hz

Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.

Explanation:

The relation that describes the pressure amplitude for a sound wave is

P_MAX = B*k*A (1)

Where the bulk modulus of the air is B = 1.30 x 10^5 Pa and the displacement amplitude of the waves produced by the machine is 1.00 μmμm.

Using (1) we can calculate k then we can use k to determine the wavelength A of the wave, and remember that λ = 2π/k.

So, substitute into (1) with 10 Pa for P_max, (1.30 x 10^5 Pa) for B and

1 x 10^-6 m for A

10 Pa = (1.30 x 10^5 Pa) x k x (1 x 10^-6 m)

k = 62.5 m^-1

We can use the following relation to calculate the wavelength

λ = 2π/k.

λ = 0.100 m

Finally, the relation between the wavelength and the frequency of a sound

wave is given by the following equation

f = v/ λ

=344/0.100 m

f =3.4*10^3 hz

Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.

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ki77a [65]

Answer:

Current flows in a resistor-capacitor circuit because of the varying electric field across the plates of a capacitor induced by an AC voltage source <em>(displacement current)</em>

Explanation:

In a capacitor, current does not flow the same way it does in a circuit, that is through conduction. This is because there is a highly resistive material in between the plates of the capacitor. Rather current flows through a phenomenon called displacement current.

Because of change in charge accumulation with time above the plates, the electric field changes causing the displacement current.

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2 years ago

The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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2 years ago

Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a

Tema [17]

Note: The diagram referred to in the question is attached here as a file.

Answer:

The magnitude of the magnetic field is $B = \frac{0.071 \mu I}{d}$

Explanation:

The magnetic field can be determined by the relationship:

$B = \frac{\mu I}{2\pi R}$ ...............(1)

Were I is the current flowing through the wires

The distance R from point 1 to m is calculated using the pythagora's theorem

$R = \sqrt{d^{2} + (2d)^{2} }$

$R = \sqrt{5d^{2} } \\R = d\sqrt{5}$

Substituting R into equation (1)

$B = \frac{\mu I}{2\pi d\sqrt{5} }$

$B = \frac{0.071 \mu I}{d}$

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2 years ago

6. Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate o

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For the answer to the question above,

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the car has to travel 1.4 miles OR 1.6 miles. </span>

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7 0

2 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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2 years ago