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Scorpion4ik [409]
3 years ago
12

A seismographic station receives S and P waves from an earthquake, separated in time by 17.3 s. Assume the waves have traveled o

ver the same path at speeds of 4.50 km/s and 7.80 km/s. Find the distance from the seismograph to the focus of the quake.
Physics
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer:

184 Km

Explanation:

given,

speed of S wave = 4.50 Km/s

speed of  P wave = 7.80 Km/s

reading time difference = 17.3 s

we know,

distance = speed x time

 time taken by s-wave

t_1 = \dfrac{x}{4.50}

time taken by the P-wave

t_2= \dfrac{x}{7.80}

now,

t₁ - t₂ = 17.3 s

\dfrac{x}{4.50}-\dfrac{x}{7.80} = 17.3 s

\dfrac{(7.80-4.50)x}{7.80\times 4.50} = 17.3

 3.3 x = 607.23

    x = 184 Km

distance of the focus from the station is 184 Km.

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Answer:

PAPER CLIPS ON NOSE OF A PAPER AIRPLANE

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Explanation:

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3 years ago
Most of the substances around you are _______
monitta

Answer:

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Explanation:

5 0
3 years ago
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Compute the torque about the origin of the gravitational force F--mgj acting on a particle of mass m located at 7-xî+ yj and sho
Andrews [41]

Answer:

Explanation:

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r = - 7x i + y j

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3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

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(a)

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E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

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(b)

As we know,

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⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

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or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch
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3 years ago
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