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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.

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A mass of 630g is hung on a spring. Using Force = mass x gravity, what is the force of the mass, acting on the spring?
hodyreva [135]

Answer:

6.174N or roughly 6.2N

Explanation:

630g=0.63kg

F=m*g=0.63*9.8=6.174N

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A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t
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I don’t know what the angle is in your diagram so I used the angle from the vertical.

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A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c
Reika [66]
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

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vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
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8 0
2 years ago
A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o
kvasek [131]

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

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Substituting:

x = 500 nm / (4 * 1.49)
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6 0
2 years ago
The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to
Taya2010 [7]

Answer:

B.   i=2.79A

C.   F=0.066N

Explanation:

A) By the right hand rule we have that

F=iL x B

F=iLBsin(α)

If the wire jump toward the observer the top pole face is the magnetic southpole.

B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

F=iLBsin(\alpha)\\\alpha=90°\\F=iLB\\i=\frac{F}{LB}=\frac{6.71*10^{-2}N}{(0.15m)(0.16T)}=2.79A

C) Now the length of the wire that feels B is

L=\frac{0.15m}{cos(10\°)}=0.152m

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

F=iLBsin(\alpha)\\F=(2.79A)(0.152m)(0.16T)(sin(80\°))=0.066N

I hope this is useful for you

regards

8 0
2 years ago
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