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3 years ago

Use the accompanying seismogram to answer which of the three types of seismic waves reached the seismograph first.

Physics

1 answer:

UkoKoshka [18]3 years ago

6 0

Answer:

Primary waves (P-waves)

Explanation:

Due to excess of the energy inside the earth when the tectonic plates begin to slide or fracture then the energy is released in the form of seismic waves, this causes the earthquake.

<u>Two types of seismic waves are generally responsible for the earth quakes:</u>

body waves
surface waves

Body waves are of two types:

Primary waves (P-waves)

These are the fastest of all the waves involved in the earth-quake which travel at a speed of 1.6 km to 8 km per second.

They can pass trough solids, liquids and gases. They arrive at the surface as an instant thud.

Secondary waves (S-waves)

They can only pass through the solids and they move slower than the P-waves.

As S-waves move, they displace the rock particles, pushing them outwards perpendicular to the wave-path that leads to the earthquake-related first rolling period.

Surface waves (L-waves/ long waves)

These waves move along the surface of the earth. They are responsible for the earthquake's carnage.

They move up and down the Earth's surface, rocking the foundations of man-made structures.

Surface waves are slowest of the three waves, which means that they are the last to arrive. So at the end of an earthquake usually comes the most powerful shaking.

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A particle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. Assume frictionless mot

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Answer:

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º, T total=2*Voy/g

Explanation:

Velocity in the Y axis:

Voy=Vo*sinα

Time to reach the maximal height :

Kinematics equation: Vfy=Voy-at

a=g*sinα ; g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

Time required to return to the starting point:

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

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3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

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<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

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