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bija089 [108]
3 years ago
6

HELP ASAP!!! Some bands hate playing in school gyms because sound waves easily reflect off the walls and floor. What could you d

o to the gym walls to improve the sound quality?
Physics
1 answer:
jonny [76]3 years ago
4 0

Answer:

1.) Covering the ceiling and walls with soft perforated boards

2.) Hanging curtains round the hall

3.) Having more opening in the wall.

Explanation:

This is due to what we called Reverberation due to poor acoustic properties.

Reverberation can be reduced by;

1.) Covering the ceiling and walls with soft perforated boards

2.) Hanging curtains round the hall

3.) Having more opening in the wall.

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As it travels through a crystal, a light wave is described by the function E(x,t)=Acos[(1.57×107)x−(2.93×1015)t]. In this expres
Drupady [299]

Answer:

Speed, v=1.86\times 10^8\ m/s

Explanation:

It is given that,

A light wave is described by the following function as :

E(x,t)=A\ cos[(1.57\times 10^7)x-(2.93\times 10^{15})t].....(1)

The general equation of wave is given by :

E=Acos(kx-\omega t)........(2)

On comparing equation (1) and (2)

k=(1.57\times 10^7)

\dfrac{2\pi}{\lambda}=(1.57\times 10^7)

\lambda=\dfrac{2\pi}{(1.57\times 10^7)}

Wavelength, \lambda=4.002\times 10^{-7}\ m

\omega=(2.93\times 10^{15})

\dfrac{2\pi}{T}=(2.93\times 10^{15})

\dfrac{1}{T}=\dfrac{(2.93\times 10^{15})}{2\pi}

Frequency, f=4.66\times 10^{14}\ Hz

Let v is the speed of the light wave. It is given by :

v=f\times \lambda

v=4.66\times 10^{14}\ Hz\times 4.002\times 10^{-7}\ m

v=1.86\times 10^8\ m/s

So, the speed of the light wave is 1.86\times 10^8\ m/s. Hence, this is the required solution.

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3 years ago
How is air resistance similar to gravity? give me two reasons
QveST [7]

Explanation:

I think this would help you. Read this and make your own answer ok.

8 0
2 years ago
135g of an unknown substance gains 9133 J of heat as it is heated from 25⁰C to 100⁰C. Using the chart below, determine the ident
telo118 [61]

Answer:

The unknown substance is Aluminum.

Explanation:

We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:

Initial temperature (T₁) = 25 ⁰C

Final temperature (T₂) = 100 ⁰C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 25

ΔT = 75 ⁰C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Change in temperature (ΔT) = 75 ⁰C

Mass of the substance (M) = 135 g

Heat (Q) gained = 9133 J

Specific heat capacity (C) of substance =?

Q = MCΔT

9133 = 135 × C × 75

9133 = 10125 × C

Divide both side by 10125

C = 9133 / 10125

C = 0.902 J/gºC

Thus, the specific heat capacity of substance is 0.902 J/gºC

Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.

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3 years ago
Your friend claims that weather predictions are just guesses. What would you say to change your friend's mind?
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To change my friends mind, after they've already said weather predictions are just guesses, I would say they are correct, but not only to they use guesses to predict what will happen, but they also, before hand, use atmospheric tools. They then predict from the data collected. They do both.
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3 years ago
In terms of volume,how do ml & cm3 relate to one another?
Keith_Richards [23]

1 milliliter = 1 cubic centimeter (cm^3)

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3 years ago
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