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Sunny_sXe [5.5K]
3 years ago
14

When earthquakes produce vibrations called waves, how do those waves travel? They ripple from the epicenter of the earthquake. T

hey move back and forth and side to side through water. They travel away from the focus of the earthquake in all directions. They move slowly and increase in intensity as they reach Earth's surface.
Physics
2 answers:
Lana71 [14]3 years ago
7 0

Answer: They travel away from the focus of the earthquake in all directions.

Explanation:

The vibrations produced by Earthquake are called seismic waves. seismic waves travel from the point where fault occurs. The maximum intensity is about the focus of the fault. These waves travel away from the focus in all directions.

Seismic waves are both transverse (S waves) and longitudinal (P waves). The P and S waves can travel through the Earth where as the surface waves travel above or near the Earth's surface.

kozerog [31]3 years ago
5 0

Answer: (C)

Explanation:  They travel away from the focus of the earthquake in all directions.

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Two identical satellites orbit the earth in stable orbits. one satellite orbits with a speed v at a distance r from the center o
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The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r

Solution:
The correct answer is C) at a distance greater than r.

In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
G \frac{Mm}{r^2}=m \frac{v^2}{r}
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed

Re-arranging the equation, we get
r= \frac{GM}{v^2}
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.
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With which part of the brain is awareness typically associated?
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A 57 kg wagon is pulled with a constant net force of 38 N. Calculate the acceleration of the wagon. F = ma
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Answer:

<h2>0.67 m/s²</h2>

Explanation:

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<h3>0.67 m/s²</h3>

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2 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
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1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
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2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

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3 years ago
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