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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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There is an electromagnetic wave traveling in the -z direction in a standard right-handed coordinate system. What is the directi
wlad13 [49]

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (<em>in this case the -z direction</em>) and the curling of the fingers point in the direction of the magnetic field  B→ (<em>in this case the +x direction</em>), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

6 0
2 years ago
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Q₁ = 572.8 * 10^{-9} C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to
photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

7 0
2 years ago
Read 2 more answers
B. Complete the table to show the effect of each change on each electric quantity. (1 point)
notka56 [123]

Answer:

Effect on electric force

Multiply one charge by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is directly proportional to the charge.

Hence if one charge is doubled, then the electric force is doubled.

Multiply distance by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is inversely proportional to the square of the distance of separation.

If the distance is doubled, F is decreased by 22. This means that the force is multiplied by 1/4.

Effect on electric potential energy

Multiply one charge by 2.

The electric potential energy is given by U=kq1q2/r

From the equation, the electric potential energy is directly proportional to the charge q.

If one charge is doubled, the electric potential energy is doubled.

Multiply distance by 2

The electric force is given by U=kq1q2/r

From the equation, the electric potential energy is inversely proportional to the distance of separation r.

If the distance is doubled, U is divided by 2. This means that the electric potential energy is multiplied by 1/2.

Effect on potential difference

Potential difference is defined as the change in electric potential energy.

Increase in the charge causes an increase in the potential difference and an increase in the distance of separation decreases the potential difference.

4 0
2 years ago
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