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Thepotemich [5.8K]
3 years ago
8

The sound wave produced from playing the A key one octave above the "Middle C" on the plano produces a sound wave with a frequen

cy of 880 Hz. The speed of sound moving through the air at room temperature is 344 m/s. What is the wavelength of the sound waves associated with the A key?
Physics
2 answers:
Vaselesa [24]3 years ago
4 0

Answer: well I think you have to take 800 Hz x 344 m/s Which should equal around 275200 ;-; and then divide it by something (I did this once but I cant remember the rest .-.)

natita [175]3 years ago
3 0

Answer:

<u>Frequency</u>- number of wave cycles that occur in a given amount of time.

<u>Pitch</u>- number of wavelengths in a given amount of time.

<u>Amplitude</u>- fluctuation or displacement of a wave from its mean value. That means how high or low they are away from the center line.

<u>Volume</u>- The perception of loudness from the intensity of a sound wave. The higher the intensity of a sound, the louder it is perceived in our ears, and the higher volume it has.

<u>Wavelength</u>- the distance between the tops of the "waves".

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Search about non-ohmic devices in 3 pages​
Tomtit [17]

Answer:

electronic diode,

Explanation:

Non-ohmic conductors are materials that do not obey ohm's law and they are electronic diode, transistors, tungsten, thermistors and vacuum tube etc.

7 0
3 years ago
A person travelled in the -x direction at a constant speed of 2.5 m/s. The person's final position is -10.0 m, and the initial p
Nostrana [21]

Answer:

(c) time required to travel = 8 sec

Explanation:

We have given the final position = -10 m on x axis

And the initial position =10 m

So total distance = 10-(-10)=20 m

The speed is given as 2.5 m/sec

We have tof ind the time required by the person to travel

Time is given by t=\frac{distance}{speed}=\frac{20}{2.5}=8 sec

So the option (c) is correct option

4 0
3 years ago
How long will it be before 82.0<br> % of an isotope with a half life of 1250<br> y decays?
Solnce55 [7]
Can you give more information
5 0
3 years ago
An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob
lorasvet [3.4K]

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N

Therefore, the force of attraction between the two charges is 3.55 N.

3 0
2 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
3 years ago
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