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lorasvet [3.4K]
2 years ago
10

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?
Physics
1 answer:
erastova [34]2 years ago
6 0

Answer:

It is somewhere in between

Explanation:

Wave length of sound from each of the  speakers = 340 / 1700 = 0.2 m = 20 cm

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

Distance between second speaker and the given sound

D = √(4² + 2²)

D = √(16 + 4)

D = √20

D = 4.472 m

Path difference = 4.472 - 4 =

0.4722 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 is of neither a perfect constructive nor a perfect destructive interference.

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kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
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The burning of fossil fuels contributes to the addition of greenhouse gases to the atmosphere. These gases trap thermal energy i
denpristay [2]

Answer:

B. there would be a global rise in temperatures

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Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

N = 7.74 rev

Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

so we have

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)

\alpha = 0.48 rad/s^2

Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

12.56 - 8.17 = 0.48 t

t = 9.14 s

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2 years ago
9ma electric current is flowing through a conducting wire , then the number of electron passing through it in 3 min is
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Answer:

1.0125 x 10^19

Explanation:

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charge passing per 3 min

Q = It

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= 1.620 C

no of electrons in charge

Q = ne

1.620 = n x 1.6 x 10 ^ -19

n. = 1.0125 x 10 ^19

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Answer: a) The Answer to the question is option a) None of it.

Explanation:

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