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dangina [55]
3 years ago
5

On the New York seismogram, the first P wave was recorded at 9:01 UTC. UTC is the international standard on which most countries

base their time. At what time (UTC) did the earthquake actually occur? Explain.
Physics
1 answer:
nevsk [136]3 years ago
4 0
What do you mean by "actually" ? ? ?

Do you mean what time was it in New York when the earthquake began ?
OR
Do you mean what time was it at the place where the earthquake occurred
when the earthquake began ?

New York is in the Eastern Time Zone.  That's five (5) zones west of UTC.

If the quake occurred between November and March, during the time
when  the US is on Standard time or "Winter" time, then the first P-wave
was recorded in New York at  4:01 AM  EST  that morning.

If the quake occurred between April and October, during the time when 
the US is on Daylight Saving time or "Summer" time, when clocks are set
1 hour ahead, then the first P-wave was recorded there at  5:01 AM  EDT 
that morning.

We have no way to calculate what the time was in the place where the earthquake occurred, because the question gives no information
regarding the location of its epicenter.  We would need to know
not only its time zone, but also the actual distance from New York,
so that we could estimate the length of time it took for the P-wave
to reach the seismograph there.
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3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
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Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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Answer:

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By Ohm's Law, 

V=IR

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R=4.8×103 Ω

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