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Tamiku [17]
3 years ago
13

A student used a yardstick to model the displacement of rock at a fault during an earthquake. The student bent the yardstick wit

hout breaking it and then let it go. What process did the student show?
A. A moment magnitude vibration
B. Elastic rebound
C. Isostasy
D. The formation of a new fault
Physics
2 answers:
jonny [76]3 years ago
7 0
I think the correct answer would be B. The process of elastic rebound is being shown by the student. It is a theory that is used to explain earthquakes. It focuses on how energy is being spread in times of earthquakes. As the rocks on the fault experiences shift and force, these rocks would be accumulating energy causing it to deform reaching the internal strength and eventually exceeding it. At that moment, a rapid motion would happen along the fault, which releases the energy, then the rocks would go back to its original shape or the undeformed state. This theory is the first theory that sufficiently was able to explain earthquakes.
dlinn [17]3 years ago
6 0

Answer:

For APEX: Elastic Rebound

Explanation:

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What is the inverse square law and how does it relate to gravity?
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Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

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4 years ago
How does changing the voltage in a circuit affect the current in the circuit
Tasya [4]
The formula for the voltage is shown below:
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7 0
3 years ago
Read 2 more answers
A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of
Lynna [10]

We have the relation

\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}

where v_{A \mid B} denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}

v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}

Let's assume the river flows South-to-North, so that

\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath

and let -90^\circ < \theta < 90^\circ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)

Then the velocity of the boat relative to the Earth is

\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath

The crossing is 153.0 m wide, so that for some time t we have

153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s

which is minimized when \theta=0^\circ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}

The boat's position \vec x at time t is

\vec x = \vec v_{B\mid E} t

so that after 30.0 s, the boat's final position on the other side of the river is

\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath

and the boat would have traveled a total distance of

\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}

3 0
2 years ago
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zloy xaker [14]

Answer:

1) A,B,F,G

2)B,D,E

3) A,C,F,G

Explanation:

8 0
3 years ago
What is the period in seconds for 10 oscillations in 5 seconds
saul85 [17]

Answer:

I think the answer is half a second cause Time period = Time taken by the pendulum/Number of oscillations made by the pendulum so 5 divided by 10 equals 0.5 which is half a second

5 0
3 years ago
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