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2 years ago

What is a method to predict an earthquake?

1 answer:

Pie2 years ago

8 0

Measurements of microtremors, slope of the land, and motion of tectonic plates predict earthquakes. But we can't predict WHEN accurately enough yet to make the prediction useful to the general public.

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Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h

nekit [7.7K]

Answer:

2.62seconds

Explanation:

Speed is defined as the ratio of the distance covered by a body with respect to time.

Speed v = Distance (s)/Time (t)

For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.

Speed v = 2 × distance (S)/Time (T)

V = 2S/t

2S = vt

Given speed of the wave = 342m/s

Distance covered = 450m

t = 2S/v

t = (2×450)/343

t = 900/343

t = 2.62seconds

It will take him 2.62seconds for him to hear his own voice echo off of the wall.

5 0

2 years ago

Read 2 more answers

Where is the Oort cloud located?

sergiy2304 [10]

Answer: C

Explanation: just did it

8 0

2 years ago

Read 2 more answers

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

2 years ago

Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?

laila [671]

Approximately 101 N air is in a column 1-cm2 in cross-section that extends from sea level to the top of the atmosphere

The basic level for determining height and depth on Earth is the sea level. The ocean's surface tends to seek the same level since it is one continuous body of water. However, the sea level is never fully level due to winds, currents, river discharges, and changes in gravity and temperature.

At the equator, the radius of the Earth at sea level is 6378.137 km (3963.191 mi). At the poles, it is 6,356.752 km (3,949.903 km), and on average, it is 6,371.001 km (3,958.756 mi). The elevation of the shoreline—the boundary between the ocean and the land—is referred to as sea level. Land that is higher than this altitude is above sea level, and land that is lower is below sea level.

To learn more about sea level please visit -
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#SPJ4

8 0

1 year ago

The line graph below shows the number of downloads of two songs after

daser333 [38]

Answer: D

1200

Explanation:

Song 1 is spotted with a cube sign.

At 3 minute, trace the spot to the vertical axis. And you will notice that it a little bit above 10.

Since it is above 10, let assume it is equal to 12.

The number of song downloaded are in hundreds. Therefore, multiply the 12 by 100

12 × 100 = 1200 downloads

Approximately, song 1 has 1200 downloads at minute 3

5 0

2 years ago