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zavuch27 [327]
3 years ago
6

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directi

on in which such pressure fluctuations travel?A) The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.B) Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.C) Pressure fluctuations travel along the direction of propagation of the sound wave.D) Propagation of energy that passes through empty spaces between the particles that comprise the mediumDoes air play a role in the propagation of the human voice from one end of a lecture hall to the other?a) yesb) no
Physics
1 answer:
Julli [10]3 years ago
3 0

Answer:

None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave

Explanation:

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Why does it takes a long time to heat a room with a high ceiling?
vlada-n [284]

Answer:

High ceilings make a room feel large and open, but they can be difficult to cool and heat. Because hot air rises, the challenge becomes trying to keep the hot air where you want it and preventing if from being wasted where you don't.

Explanation:

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3 years ago
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g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
4 years ago
Change in speed over a given period of time is
Dvinal [7]

Explanation:

Acceleration is the change in speed over a given time period

8 0
4 years ago
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Four different resistors have various amounts of electric current flowing through them. Given the values of current I and resist
Artemon [7]

You did not provide the options. However, the options are

I = 6.0, R= 4.0 ohms

I = 9.0, R= 2.0ohms

I = 3.0, R= 2.0ohms

I = 8.0, R= 8.0 ohms

Answer:

The order of  the resistors from the highest to the lowest is:

I = 8.0, R= 8.0 ohms

I = 6.0, R= 4.0 ohms

I = 9.0, R= 2.0ohms

I = 3.0, R= 2.0 ohms

Explanation:

ohm's law states that voltage across a conductor is directly proportional to the current flowing through it.  V = IR

Based on this formula, the voltages in each of the resistors are calculated below from the highest to the lowest

  • For I = 8.0, R= 8.0 ohms

       V = 8 * 8 =64 volts

  •  For I = 6.0, R= 4.0 ohms

      V = 6 * 4 =24 volts

  • For I = 9.0, R= 2.0 ohms

       V = 9 * 2 =18 volts

  • For I = 3.0, R= 2.0 ohms

       V = 3 * 2 =6 volts

3 0
3 years ago
Define the term overload.
ryzh [129]

Answer:

Verb. Load with too great a burden or cargo.

"Both boats were overloaded and low in the water"

Noun. An excessive load or amount.

"an overload of stress"

Explanation:

Similar words are strain, excess, and overburden.

Have a good day and stay safe!

4 0
3 years ago
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