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const2013 [10]
2 years ago
5

When earthquakes occur, seismic waves are sent through the Earth. The two most common types of seismic waves are P waves and S w

aves. The graph below shows how P waves and S waves can travel through the Earth.
Based on the graph, what is true about P waves and S waves?
A. They both travel faster than 7,000 miles per minute
B. They travel at different speeds through the Earth.
C. They both travel slower than 1,000 miles per hour.
D. They travel at the same speed through the Earth.
PLEASE HELP!!

Physics
1 answer:
dexar [7]2 years ago
4 0

Answer:

B. They travel at different speeds through the Earth.

Explanation:

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Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, m_1=1000\ kg

Initial velocity of first car is, u_1=30\vec{j}\ m/s

Mass of second car is, m_2=1250\ kg

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, u_{2y}=0.

The magnitude of final velocity is given as:

|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

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Answer:

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B) a star that is close to the south celestial pole: a circumpolar star could be close to the south celestial pole, but this answer is omitting the north celestial pole.

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