Ask question

Ask question

All categories

3 years ago

14

Two sources of in-phase coherent sound are located at the points (0 m, 2 m) and (-2 m, 0 m). An observer at the origin hears con

structive interference because she is equidistant from both sources. However, if she moves in the +x direction, she hears destructive interference for the first time when she reaches the point (0.6 m, 0 m). What is the frequency of the sound that the sources are emitting? a) 476 Hz b) 391 Hz c) 335 Hz d) 295 Hz e) 244 Hz

1 answer:

Korolek [52]3 years ago

3 0

Answer:

c) 335 Hz

Explanation:

Given that the two sources:

S₁ ≡ (0 m, 2 m)

S₂ ≡ (-2 m, 0 m)

The lady hears the destructive interference for the first time when she reaches the point (0.6 m, 0 m).

So, the path difference = Δx = |S₂P - S₁P|

$S_2P=\sqrt {(-2-0.6)^2+(0-0)^2}=2.6\ m$

$S_1P=\sqrt {(0-0.6)^2+(2-0)^2}=2.0881\ m$

So,

Δx = |2.6 - 2.0881| = 0.5119 m

For, first destructive interference, Δx = λ / 2

So,

Wavelength = 1.0238 m

Also,

c = ν×λ

Where,

c is the speed of sound wave having value as 346 m/s

ν is the frequency

λ is the wavelength

So,

<u>ν = c / λ = 346 m/s / 1.0238 m ≈ 335 Hz</u>

You might be interested in

Impact of electricity in the society

Travka [436]

Answer:

<h2>Electricity has many uses in our day to day life. It is used for lighting rooms, working fans and domestic appliances like using electric stoves, A/C and more. All these provide comfort to people. In factories, large machines are worked with the help of electricity.</h2>

5 0

2 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago

(01.01 LC) Two boxes are 8 cm apart. Which of the following should Janet do to

Trava [24]

Place the boxes 10cm apart, if they are closer that concentrates the mass of the boxes more

4 0

3 years ago

Read 2 more answers

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

3 years ago

An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp

Natali5045456 [20]

Answer:

<em>h = 20 m</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or $9.8 m/s^2$ .

The weight of an object of mass m is:

W = m.g

Thus, the GPE is:

U=W.h

Solving for h:

$\displaystyle h=\frac{U}{W}$

The weight of the owl is W=40 N and its GPE is U=800 J.

$\displaystyle h=\frac{800}{40}=20$

h = 20 m

3 0

3 years ago