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Fofino [41]
2 years ago
5

Two speakers emit sound waves with frequency 4.55 kHz. They are driven by the same oscillator so that they are in phase with eac

h other. I place the speakers side-by-side, and I stand across the room from them. If someone moves one of the speakers towards me, I hear the total intensity drop and then rise again. How far had they moved the speaker at the point where I heard a minimum intensity due to destructive interference of the sound waves from the two speakers? The speed of sound is 343 m/s
Physics
1 answer:
sergiy2304 [10]2 years ago
4 0

Answer:

d=3.76 cm

Explanation:

Given that frequency ,f= 4.55 KHz

Speed of sound = 343 m/s

We know that for sound

C= f λ

Where C is the velocity of sound

f is the frequency

λ is the wavelength

So now by putting the values

C= f λ

343= 4550 λ

λ=7.53 cm

We know that for destructive phase the difference in phase is π so the difference in distance will be λ/2.

So distance ,d= λ/2.

d=7.53/2 cm

d=3.76 cm

So they will move at a distance of 3.76 cm to hear minimum intensity of sound.

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Originally there must been

1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start

% = 2.25 / 150 = 1.5 %      of 235 U left

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What is the net charge of a copper atom if it gains 2 electrons?
Alex_Xolod [135]
If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.
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3 years ago
What is Potentiometer ​
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an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

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2 years ago
An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet
zimovet [89]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

        F = 1.6 10⁻¹⁹  1.344 10⁸ 0.4

       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

5 0
2 years ago
A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
balandron [24]

Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

Hence:

\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

In other words, the angle of refraction in the glass would be approximately 28^{\circ}.

7 0
2 years ago
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