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oksian1 [2.3K]
4 years ago
6

How do you measure the amplitude of a longitudinal wave?

Physics
1 answer:
sp2606 [1]4 years ago
3 0

You do it by the pressure difference it causes in the medium. Even in transversal waves, you usually need different units than meters because the wave has electric and magnetic vectors and not really any height to speak of.

you measure the amount of excursion from the equilibrium (no wave) level of whatever parameter is fluctuating.


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The lowest note on the piano has the longest string to vibrate. How does the lowest note compare if you consider the column of a
SashulF [63]
The piano strings<span> for </span>low notes<span> are heavier, </span>have<span> more inertia, and </span>vibrate<span> at a lower frequency a lower pitch than lighter </span>strings<span> of the same </span>string<span> tension. Loudness involves how hard the keys are struck, which affects the amplitudes of the </span>vibrating strings<span>. the touch sensitivity of the </span>piano<span> distinguishes it from earlier.



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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Gnesinka [82]

Answer:

a).a_p=-2.39x10^{-12} rad/s^2

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Explanation:

a).

The acceleration for definition is the derive of the velocity so:

a_p=\frac{dw}{dt}

w=\frac{2\pi}{t}

a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}

dT=0.0808s

dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s

Replacing

a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2

b).

If the pulsar will continue to decelerate at this rate, it will  stop rotating at time:

t=\frac{w}{a_p}

w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s

t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s

t=1016298.8 years

c).

582 years ago to 2019

1437

T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s

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I think that it might be B?
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