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aev [14]
2 years ago
6

Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting

1700-Hz sound waves into a room where the speed of sound is 340 m/s. Consider a point 4.0 m in front of speaker 1, which lies along a line from speaker 1, that is perpendicular to a line between the two speakers. Is this a point of maximum constructive interference, a point of perfect destructive interference, or something in between? Compute the path-length difference ? What is the wavelength of the sound waves emitted by the speakers?
Physics
1 answer:
horsena [70]2 years ago
7 0

Answer:

Explanation:

Wave length of sound from each of the  speakers = 340 / 1700 = .2 m = 20 cm

Distance between first speaker and the given point = 4 m.

Distance between second speaker and the given sound

= √ 4² + 2² = √16 +4 = √20 = 4.472 m

Path difference = 4.472 - 4 = .4722 m.

Path difference / wave length = 0.4772 / 0.2 = 2.386

This is a fractional integer which is neither an odd nor an  even multiple of half wavelength. Hence this point of neither a perfect constructive nor a perfect destructive interference.

You might be interested in
An ambulance driving 35.0 m/s emits a sound wave with a wavelength of 80.0 centimeters. As it drives away from a hospital, which
katen-ka-za [31]

Apparent frequency heard by the staff: 389 Hz

Explanation:

The phenomenon described in this situation is called Doppler effect.

Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:

- The observer perceives a higher frequency if the source is moving towards them

- The observer perceives a lower frequency if the source is moving away from them

The formula to calculate the apparent frequency in the Doppler effect is

f'=\frac{v\pm v_o}{v\pm v_s}f

where

f is the proper frequency

f' is the apparent frequency

v is the speed of the wave

v_o is the velocity of the observer (positive if they are moving towards the source, negative if moving away)

v_s is the velocity of the source (positive if it is moving away, negative if moving towards the observer)

First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:

v = 343 m/s (speed of sound wave)

\lambda=80 cm = 0.80 m (wavelength)

So the proper frequency is

f=\frac{v}{\lambda}=\frac{343}{0.80}=429 Hz

Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:

v_s = +35.0 m/s (velocity of the ambulance)

v_o = 0 (velocity of the staff)

Substituting,

f'=\frac{343+0}{343+35}(429)=389 Hz

Learn more about frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0
1 year ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
1 year ago
A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.
Bumek [7]

Answer:5 rad/s^2

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

Torque =I\alpha =F\cdot r

where \alpha =angular\ acceleration

I\alpha =F\cdot r

0.02\cdot \alpha =1\cdot 0.1

\alpha =5 rad/s^2    

5 0
2 years ago
An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

4 0
1 year ago
an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and
natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

\Delta p=m(v-u)\\\\\Delta p=mv

The change in momentum is equal to the impulse delivered. So,

J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.                      

4 0
1 year ago
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