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3 years ago

wink is the seismic station closest to the earthquake. describe two evidences from the seismogram that demonstrates this?

Physics

1 answer:

kotykmax [81]3 years ago

5 0

Well, we don't know if it's the CLOSEST one, but it's closer than the other ones on this print-out.

There are actually three clues to this in the seismograms:

1). The waves from the tremor reach the Wink station first ... the P wave gets there 50 seconds before Chicago and 3 whole minutes before Tampa.

2). They are stronger (greater amplitude) at Wink than at Chicago or Tampa.

3). The delay between the P wave and the S wave is the smallest at Wink. This delay between the two waves grows as the detector is farther from the quake.

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4 years ago

Line segment Q R , Line segment R S and Line segment S Q are midsegments of Î"WXY. Triangle R Q S is inside triangle X Y W. Poin

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The perimeter of ΔWXY is : ( D ) 14.5 cm

Calculating the perimeter of ΔWXY

QR = WY / 2

RS = XW / 2

QS = XY / 2

Given that : QR = 2.93 cm , RS = 2.04 cm, QS = 2.28 cm

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3 years ago

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Solnce55 [7]

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3 years ago

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A wheel rotates about a fixed axis with an initial angular velocity of 24 rad/s. During a 4 s interval the angular velocity decr

netineya [11]

Answer:

$\theta=76\ rad$

Explanation:

Hoven that,

Initial angular velocity of the wheel = 24 rad/s

Final angular velocity = 14 m/s

Time, t = 4 s

We need to find how many radians does the wheel turn through during the 4 s interval. Let the displacement is $\theta$ . Using second equation of rotational kinematics to find it such that,

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

Where

$\alpha$ is angular acceleration

$\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{14-24}{4}\\\\\alpha =-2.5\ rad/s^2$

So,

$\theta=24\times 4+\dfrac{1}{2}\times (-2.5)\times 4^2\\\\\theta=76\ rad$

So, it will turn 76 radian during the 4 s interval.

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3 years ago