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2 years ago

13

A shear wave (S wave) is a type of seismic _________ that shakes the ground back and forth perpendicular to the direction the wa

ve is moving.

1 answer:

Zepler [3.9K]2 years ago

7 0

Body waves

Explanation:

A shear wave(S-wave) is a type of seismic body waves that shakes the ground back and forth perpendicular to the direction the wave is moving.

Seismic waves are elastic waves usually generated when there is a disturbance within the earth.
There are two types of seismic waves:

Surface waves

Body waves

Body waves travel within the earth and they cause disturbances there. P and S waves are the two types of body waves that we have.
Surface waves travels on the earth surface. They are the love and rayleigh waves. They are the ones that cause destruction on the earth surface during an earthquake.

Learn more:

Earthquake brainly.com/question/6520403

#learnwithBrainly

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How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?

podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

$\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J$

Final velocity = 40 m/s

$\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J$

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

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2 years ago

Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh

guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

$V = i R$

$i = \frac{V}{R}$

$i = \frac{120}{160}$

$i = \frac{3}{4} A$

now the charge flow through it is given as

$Q = i t$

total time is t = 8 hours

$Q = \frac{3}{4}(8 \times 60 \times 60)$

$Q = 21600 C$

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2 years ago

An aircraft on it's take-off run has a steady acceleration of 3m/s^2. How much velocity does it gain 10 seconds?

cluponka [151]

Acceleration measures how fast the speed changes over time.

So, over 10s, the aircraft's speed changes by 3 $\frac{m}{ s^{2} }$ * 10s

So, the change is 30 $\frac{m}{s}$

Note that the one of the second units cancels out in the answer.

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1 year ago

Read 2 more answers

A wagon full of manure accidentally rolls down a driveway for 5.0m while a person pushes against the wagon with a force of 420 N

Cerrena [4.2K]

Answer:

2100 J

Explanation:

Parameters given:

Force acting on the object, F = 420 N

Distance moved by object, d = 5m

The change in kinetic energy of an object is equal to the work done by a force acting on the object:

W = F * d

∆KE = F * d

∆KE = 420 * 5

∆KE = 2100 J

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2 years ago

Read 2 more answers

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

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2 years ago