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2 years ago

11

Two identical speakers are 3.50 m

1 answer:

DiKsa [7]2 years ago

6 0

Answer:

The difference in distance from the speakers is 5.2 - 3.5 = 1.7 m

The listener would be 1/2 wavelength out of phase with the speakers

1/2 y = 1.7 m where y is the wavelength

y = 3.4 m the required wavelength

f = v / y = 340 m/s / 3.4 m = 100 / sec lowest frequency

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that

Harrizon [31]

Answer:

The time taken is $t = 40007 sec$

Explanation:

From the question we are told that

The diameter of the egg is $d_e = 5.5cm = \frac{5.5}{100} = 5.5*10^{-2}m$

The initial temperature of egg the $T_e = 4.3^{o}C$

The temperature of the boiling water $T_b = 100^oC$

The heat transfer coefficient is $H = 800 W/m^2 \cdot K$

The final temperature is $T_e_f = 74^oC$

The thermal conductivity of water is $k = 0.607 W/m^oC$

The diffusivity of the egg $\alpha = 0.146 * 10^{-6} m^2 /s$

Using one term approximation

We have the

$\frac{T_e_f - T_b}{T_e - T_b} = Ae^{-\lambda ^2 \tau}$

The radius is $r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m$ Note that this radius is approximation to that of a real egg

Now we need to obtain the Biot number which help indicate the value of $A \ and \ \lambda$ to use in the above equation

The Biot number is mathematically represented as

$Bi = \frac{H r}{k}$

Substituting values

$Bi = \frac{800 * 2.75 *10^{-2}}{0.607}$

$= 36.24$

So for this value which greater than 0.1 the coefficient $\lambda_1 \ and \ A_1$ is

$\lambda = 3.06632$

$A = 1.9942$

Substituting this into equation 1 we have

$\frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-9.383 \tau}$

$0.13624 = e^{-9.383 \tau}$

Taking natural log of both sides

$-1.993 = -9.383\ \tau$

$\tau = 0.2124$

The time required for the egg to be cooked is mathematically represented as

$t = \frac{\tau r^2}{\alpha }$

substituting value is

$= \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}$

$t = 40007 sec$

8 0

3 years ago

Hey if your answering this can you just answer the number 14 I already got the rest<br> thanks

Scrat [10]

Layer a is the answer

7 0

3 years ago

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The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?

Monica [59]

Hi there!

$\large\boxed{ -6.36m/s}$

Use the equation:

$v_{2} = -\frac{m_{1}}{m_{2}} v_{1}$

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

$v_{2} = -\frac{0.048}{2.95} *391$

$v_{2} = -6.36m/s$

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3 years ago

An echo is an example of sound wave A) diffraction. B) interference. C) reflection. D) refraction.

DedPeter [7]

It is an example of reflection of wave. C.

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3 years ago

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exis [7]

Answer:

5 years worth of work (aka all of the homework i currently have)

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3 years ago