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3 years ago

What is the main evidence for the structure of the Earth that occur from the generation of an earthquake?

Physics

2 answers:

Alik [6]3 years ago

4 0

Answer;

-Waves

Explanation;

-The main evidence for the structure of the Earth comes from studying waves generated during an earthquake. When stress builds up and rocks at depth break, then elastic waves, just like sound waves are sent from the point of rupture. These waves speed out and move through the earth.

-When seismometers are put on the surface of the Earth, these waves can be measured. In particular, if we have enough seismic stations and the quake is big enough, then we can take all the travel time information and invert it. That means we can obtain a graph that shows the density of rock as a function of depth.

-Once we know the density we can estimate what the actual materials are. We find that the earth has a small inner core made out of iron surrounded by a larger outer liquid iron core and that is surrounded by a rocky mantle and rocky crust.

xeze [42]3 years ago

3 0

I believe that its waves

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A 40,000 kg subway train is brought to a stop from a speed of 0.700 m/s in 0.250 m by a large spring bumper at the end of its tr

lozanna [386]

Given,

The mass of the train, m=40000 kg

The initial velocity of the train, u=0.700 m/s

The compression in the spring bumper that stopped the train, x=0.250 m

The final velocity of the train, v=0 m/s

From the equation of motion,

$v^2-u^2=2ax$

Where a is the acceleration of the train.

On substituting the known values,

$\begin{gathered} 0-0.700^2=2a\times0.250 \\ \Rightarrow a=\frac{-0.700^2}{2\times0.25} \\ =-0.98\text{ m/s}^2 \end{gathered}$

The magnitude of the force applied by the train will be equal to the magnitude of the restoring force of the spring.

Therefore,

$\begin{gathered} m|a|=kx \\ \Rightarrow k=\frac{m|a|}{x} \end{gathered}$

Where k is the spring constant of the spring.

On substituting the known values,

$\begin{gathered} k=\frac{40000\times0.98}{0.250} \\ =156800\text{ N/m} \end{gathered}$

Therefore the spring constant of the spring is 156800 N/m

Thus the correct answer is option C.

5 0

9 months ago

An undamped spring-mass system contains a mass that weighs and a spring with spring constant . It is suddenly set in motion at b

balandron [24]

Answer:

Explanation:

When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing

motion is

Mx''+ βx' + kx = F(t).

Note for an undamped system

β=0,

Then, the differential equation becomes

Mx'' + kx = F(t).

The force is in the form

F=Fo•Sinωo•t

Let solved for the homogeneous or complementary solution, I.e f(t) = 0

Using D operator

MD² + k = 0

MD²=-k

D²=-k/M

Then, D= ±√(-k/m)

D=±√(k/m) •i

So we have a complex root

Therefore, the solution is

x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]

This is simple harmonic motion that once again we prefer to write in the form

x(t) = A•Sin[ √(k/M)t + φ]

Where A=√(C1²+C2²)

and angle φ is defined by the equations

sin φ = C1/A and cos φ = C2/A.

Quantity √(k/M), often denoted by ω, is called the angular frequency.

This is called the natural frequency (ωn) of the system

ωn=√(k/M)

ωn²= k/M

Now, for particular solution

Xp=DSinωo•t

Xp' = Dωo•Cosωo•t

Xp"=-Dωo²•Sinωo•t

Now substituting this into

Mx'' + kx = F(t).

M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t

Now, let solve for D

D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t

D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)

D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)

D=Fo / (-Mωo²+k)

D=Fo / (k-Mωo²)

Divide through by k

D=Fo/k ÷ (1 -Mωo²/k)

Note from above

ωn²= k/M

Therefore,

D=Fo/k ÷ (1-ωo²/ωn²)

D=Fo/k ÷ [1-(ωo/ωn)²]

Then,

Xp=DSinωo•t

Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t

Then the general solution is the sum of the homogeneous solution and particular solution

Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]

Check attachment for the graph of homogeneous, particular and general solution.

Also, check for better way of writing the equations.

8 0

3 years ago

If the roller coaster car in the above problem were moving with twice the speed(20m/s), then what would be its new kinetic energ

Diano4ka-milaya [45]

Answer:

KE = 1.05 x105 Joules

Explanation:

KE = 4 * (1.04653 x 105 J) = 4.19 x 105 Joules.

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2 years ago

A simple pendulum on the surface of Earth is found to undergo 15.0 complete small‑amplitude oscillations in 8.75 s. Find the pen

FinnZ [79.3K]

Answer:

Length of the Pendulum is 14.4 cm

Explanation:

Time period of simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g} } \\$

time taken for one oscillation is

$T=\frac{time taken}{oscillations} \\T=\frac{8.75}{15} \\T=0.583 s$

Pendulum length on the earth surface is

$L=\frac{T^2g}{4\pi^2 }\\L=\frac{0.583\times 9.8}{4\pi^2 } \\L=0.144 m$

Length of the Pendulum is 14.4 cm

6 0

3 years ago

What is a good definition of acceleration

lilavasa [31]

Rate of change of velocity is acceleration

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3 years ago