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2 years ago

8

Suppose you are designing an amplifier and loudspeaker system to use at a rock concert. You want to make it as loud as possible.

How can you design the system to maximize the volume? Explain your answer?
Hi, I’m very confused by this question. Is it referring to the right hand rule to predict where the force and current will face?

Thanks for helping!

1 answer:

OverLord2011 [107]2 years ago

4 0

Answer is given below

Explanation:

Audio power amplifiers are found in all types of sound systems, including sound reinforcement, public address and home audio systems, as well as musical instrument amplifiers such as guitar amplifiers.
This is the last electronic step in the general audio playback series before sending the signal to the loudspeaker. So when we want maximum volume or loud sound, we have to get it with maximum output and high input and low output impedance

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The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,

KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

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2 years ago

Caelyn wanted to find out what shampoo made her hair the shiniest . Everyday she washed her hair with different shampoos and the

Arte-miy333 [17]

Answer:

IV: type of shampoo used

DV: what shampoo made her hair the skinniest

control: water

Constant: rated from scale from 1-10

Explanation:

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2 years ago

An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of

Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 / .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B) Rest length of the rocket = length of platform = 90 m

C ) length of platform as viewed by moving observer =

$\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }$

= $\frac{90}{\sqrt{1-\frac{0.92}{1 } } }$

= 321 m

D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

$31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }$

8.75 x 10⁻⁸ s .

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2 years ago

A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

2 years ago

You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m

Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

$v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y$

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

$|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}$ ≈ $6.32 m/s$

And the angle with respect to the east should be with:

$arctan(\frac{2}{6} )=18.43 \°$

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2 years ago