What is the temperature of Neptune in Fahrenheit

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved

Alina [70]

Answer:

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let $d$ represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

the line joining the microphone and the center of the two speakers, and
the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point $P_0$ would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$ .

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$ .

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$ .

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$ .

Calculate $\theta$ for each of these path differences:

$\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}$ .

In each of these case, the distance between the microphone and $P_0$ would be $9.4\, \tan(\theta)$ . Therefore:

At the first minimum, the distance from $P_0$ is approximately $1.1\; \rm m$ .
At the first maximum, the distance from $P_0$ is approximately $2.2\; \rm m$ .
At the second minimum, the distance from $P_0$ is approximately $3.4\; \rm m$ .
At the second maximum, the distance from $P_0$ is approximately $4.7\;\rm m$ .

6 0

4 years ago

rank the following media from fastest to slowest for a light wave to travel through 1 being fastest 4 being slowest ( a. gas, b.

Arturiano [62]

The answer should be B. Solid

Hope it helped

3 0

3 years ago

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

Tju [1.3M]

Answer:

$_T}=24.57Nm$

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

$_T } = Fr$

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

$_T}=(0.795)*(30.9)$

$_T}=24.57Nm$

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

$_T}=I \omega$

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

$I = mr^2$

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

$_T}=I \omega$

Making angular acceleration the subject of the formula; we have;

$\omega = \frac{_T}{I}$

ω = $\frac{24.57}{707.51}$

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a = ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

5 0

3 years ago

An ellipse has two focal points. One of the focal points is the _____. earth, moon, or Sun

Rasek [7]

An ellipse has two focal points. One of the focal points is the Sun.

Because they planets move faster when they are around the sun.

6 0

3 years ago

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What is a physical property of giant covalent structures?

pav-90 [236]

Very high melting points - Substances with giant covalent structures have very highmelting points, because a lot of strong covalent bonds must be broken. Graphite, for example, has a melting point of more than 3,600ºC. Variable conductivity - Diamond does not conduct electricity.

5 0

3 years ago

Read 2 more answers