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iVinArrow [24]
3 years ago
7

If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the

pitch of the sound, what affect will this have on wavelength
Physics
1 answer:
Viktor [21]3 years ago
6 0

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

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A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie
-BARSIC- [3]

a) 3.27\cdot 10^{-3} J

b) 11.60\cdot 10^{-3} J

c) 8.33\cdot 10^{-3} J

Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

C=13.5\cdot 10^{-6}F is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

8 0
3 years ago
What is the unit of G in the F=Gm1m2/r^2​
kobusy [5.1K]


G
has the SI units
m
3
k
g
⋅
s
2
6 0
3 years ago
Read 2 more answers
A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respecti
Morgarella [4.7K]

Answer:

The value of the effective (rms) voltage of the applied source in the circuit is 132 V

Explanation:

Given;

effective (rms) voltage of the resistor, V_R = 65 V

effective (rms) voltage of the inductor, V_L = 140 V

effective (rms) voltage of the capacitor, V_C = 80 V

Determine the value of the effective (rms) voltage of the applied source in the circuit;

V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V

Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.

6 0
4 years ago
Minnie sota hits the end of a bar 1. 2 m long with a hammer. Sketch the standing wave on the bar for the following situations. T
ella [17]

(a) The wavelength of the wave for the fundamental mode is 2.4 m.

(b) The fundamental frequency of the wave is 2,708.33 Hz.

<h3>Wavelength of the wave for fundamental mode</h3>

The wavelength of the wave for the fundamental mode is calculated as follows;

Node to Node, N → N = λ/2

L = λ/2

λ = 2L

λ = 2 x 1.2

λ = 2.4 m

<h3>Fundamental frequency of the wave</h3>

The fundamental frequency of the wave is calculated as follows;

f = v/λ

f₀ = v/2L

f₀ = (6500)/(2 x 1.2)

f₀ = 2,708.33 Hz

The diagram of the wave for the fundamental mode is in the image uploaded.

Learn more about wavelength here: brainly.com/question/10728818

5 0
2 years ago
Plz help me with this ^^^
gogolik [260]
Assuming all the landmasses continue to move towards a sort of epicenter, large mountain ranges would appear in Africa and Europe.
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4 years ago
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