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aleksklad [387]
3 years ago
8

Your friend turns the volume up on their speaker system, so that the sound intensity is 4 times greater than it was beforehand.

By what factor has the energy output of the speaker system increased?
Physics
1 answer:
erica [24]3 years ago
4 0

Answer:

4

Explanation:

We know that intensity I = P/A where P = power and A = area through which the power passes through.

Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.

So, I = P/A and I₀ = P₀/A

Now, if I = 4I₀,

P/A = 4P₀/A

P = 4P₀

Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t

Substituting P and P₀ into the equation, we have

P = 4P₀

E/t = 4E₀/t

E = 4E₀

Since the energy is four times the initial energy, the energy output increases by a factor of 4.

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I would say its a deep ocean trench

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This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim

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3 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
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A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
Like his parents, Thomas is short, stocky, and heavyset. Although he exercises regularly and eats a well-balanced meal, his body
Viktor [21]

Answer:

Genetics?

Explanation:

8 0
3 years ago
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Things you can do with an environmental engineering major and a broadcasting minor?
Vika [28.1K]

Answer:

With an  Environmental  Engineering  and a broadcasting minor

You can work as an On Air  personality that host  programs that provide your audience with  documentaries about the environments and project carried out by Environmental Engineer

and also you can work as a journalist that explore the world making research that will preserve the environment  and leveraging the media as a broadcaster to provide this research findings as a video for you audience  

Explanation:

In order to get a better understanding let define some terms

Environmental Engineer :

Environmental engineers resolve and help prevent environmental problems. They work in many areas, including air pollution control, industrial hygiene, toxic materials control, and land management. The duties of an environmental engineer range from planning and designing an effective waste treatment plant to studying the effects of acid rain on a particular area. An environmental engineer is sometimes required to work outdoors, though most of her work is done in a laboratory or office setting. Career opportunities for environmental engineers exist in consulting, research, corporate, and government positions.

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7 0
3 years ago
Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}

m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

 = 221.79  

 = 221 (approx.)

hence no of bright fringe

M = m + 1

   = 221 +1

M = 222 fringes

6 0
3 years ago
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