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3 years ago

Four seismometers stations are the minimum needed to calculate the epicenter of an earthquake.

Physics

1 answer:

Gelneren [198K]3 years ago

4 0

<u>Answer:</u>

False, Four seismometer statitons are no required to find the earthquake's epicenter.

<u>Explanation:</u>

The least number of seismometer stations sufficient for finding the epicenter of earthquake is three. Epicenter is point on earth's crust from where the seismic rupture gets originated. Using the distances to the epicenter from atleast three seismometers the epicenter of earthquake can be determined. This distance is used as the radius and 3 circles are drawn. The point at which the 3 circles intersect is called as epicenter of the earthquake. With one seismometer station, we can determine only the distance to the epicenter. With two seismometer station, measurements are ambiguous. Using more than three seismometer stations produces redundant information. Hence three seismometer is sufficient to determine the epicenter of earthquake.

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Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

= 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

= $0.5 x 25 x 5.6^{2}$ = 392 J

change in KE (ΔKE) = initial KE - final KE
ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$
when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

\frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

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3 years ago

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Andrej [43]

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A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

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$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

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3 years ago

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Lana71 [14]

Answer:

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