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sesenic [268]
3 years ago
7

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

ed by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s.What is the closest you can be to speaker B and be at a point of perfectly destructive interference?
Physics
2 answers:
Tamiku [17]3 years ago
8 0

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of  A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am  1 m away from B , the path difference will be

8 - 1 = 7 m  which is  odd multiple of 1 or λ /2

So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

Zolol [24]3 years ago
3 0

Answer:

1m.

Explanation:

Frequency = f = 172 Hz

Speed = V = 344 m/s

For the destructive interference,

Wavelength = λ = V/f = 344/172 = 2m

Path difference = d2 – d1 = n λ/2

Where “n” is an odd integer.

If you are 1m away from the loudspeaker B then the path difference will be,

Path difference = d2 - d1 = 8 – 1 = 7m  

Which is odd multiple of λ/2 and also condition for destructive interference. Hence, to hear the perfect destructive interference you have to stay 1m close to loudspeaker B.  

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Arada [10]

Answer : The final temperature is, 25.0^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ice = 2.09J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ice = 50 g

m_2 = mass of water = 200 g

T_f = final temperature = ?

T_1 = initial temperature of ice = -15^oC

T_2 = initial temperature of water = 30^oC

Now put all the given values in the above formula, we get:

50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC

T_f=25.0^oC

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5 0
3 years ago
A cold front moves through and the temperature drops by 20 degrees. in which temperature scale would this 20 degree change repre
nalin [4]
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In the Farenheit scale, each degree is one part of 180 degrees. This is because in this scale the difference between the boiling and freezind temperatures are 212 ° - 32 ° = 180°, so one degree Farenheti is one part of 180.

That means that 1 °C is a larger amount than 1 °C, so 20°C is a larger amount than 20°F.

Conclusion: 20 degree change represents a larger change in Celsius scale.

4 0
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Answer:

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Explanation:

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7 0
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Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

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x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

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General solution for this differential eq is;

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Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

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from (2) we have;

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put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

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Answer:

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