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2 years ago

Which statements concerning how geologists locate an earthquake’s epicenter are accurate? Check all that apply.

Physics

2 answers:

Nesterboy [21]2 years ago

8 0

-- Geologists use seismic waves to locate the center of an earthquake.

-- Geologists use data from three or more data stations to determine the location of the epicenter.

-- A seismograph collects the data that makes it possible to measure the time-difference between the arrivals of P waves and S waves.

Brut [27]2 years ago

7 0

Answer:

Geologists use seismic waves to locate the center of an earthquake.

Geologists use data from three or more data stations to determine the location of the epicenter.

A seismograph measures the difference between the arrivals of P waves and S waves.

Explanation:

Seismic waves refer to the energy within the Earth due to sudden movement and explosions. Basically, they are the energy traveling within the Earth, and this energy cause movement, an excess of it will cause major movements, causing damage in the surface.

So, the epicenter of a earthquake is tracked using seismic waves, but this process more complex that we thought, because geologists need to track the epicenter using other data stations which offer real time information about seismic waves. Using all this data, geologist can find the epicenter faster and accurately.

There are different types of seismic waves, there's the Primary Waves or P-Waves, they are called primary because they are the first to perceive by the seismic station, they push and pull constantly, like a spring. Another type of seismic waves is The Secondary Wave or S-Waves, this is slower than the first one, this type of waves can moves mass of rocks up and down, constantly perpendicular to its direction. These two types are called body waves, and they are used by geologist to locate an earthquake.

Lastly, we have the surface waves, which only travel through the crust. In this category we have Love Waves which are the fastest, moving side to side. Also, we have the Rayleigh Waves, which are waves that roll the ground like ocean waves roll the water.

Therefore, based on these definitions, there are three correct answers, first, third and last one.

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Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

2 years ago

Ariel dropped a golf ball from her second story window. The ball starts from rest and hits the sidewalk 3.5 s later with a veloc

Aleks [24]

Answer:

By using the acceleration formula,

$a = \frac{v - u}{t}$

$a = \frac{14.7 - 0}{3.5}$

$a = 4.2m \: s ^{ - 2}$

4 0

2 years ago

A wire of 1mm diameter and 1m long fixed at one end is stretched by 0.01mm when a lend of 10 kg is attached to its free end.calc

Otrada [13]

Answer:

E = 1.25×10¹³ N/m²

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Given:

F = 10 kg × 9.8 m/s² = 98 N

L = 1 m

dL = 10⁻⁵ m

A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²

Solve:

E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)

E = 1.25×10¹³ N/m²

Round as needed.

5 0

2 years ago

An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

2 years ago

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit

Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle $\theta_{a}$ , since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are $n_{red} = 2.410$ for red light and $n_{blue} = 2.450$ for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has $n_{air} = 1.000$ .

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

$\theta_{crit} = 24.09^{0}$

Explanation:

Only the blue refracted ray is related to the critical angle in this question

$n_{air} = 1.000$

$n_{blue} = 2.450$

The relationship between the critical angle( $\theta_{crit}$ ), $n_{air}$ and $n_{blue}$ can be given as $sin \theta_{crit} = \frac{n_{air} }{n_{blue} }$

$sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\ \theta_{crit} = 24.09^{0}$

6 0

2 years ago