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Aleks04 [339]
3 years ago
15

If you wanted to live where the chances of a destructive earthquake were small, would you pick a location near a fault zone, nea

r a mid ocean ridge, near a subduction zone, or on a volcanic island such as Hawaii? What are the relative risks of earthquakes at each of these locations?
Physics
1 answer:
exis [7]3 years ago
5 0

Answer: A volcanic island arc like Hawaii.

Explanation: The Hawaiian island is the most safest place for the people to live in comparison to the other given options.

The plate tectonic movement generates energy due to collision, resulting in sudden release of energy, and seismic waves are produced that propagates and causes earthquake.

Due to the constant collision between the plates, different places at different time experiences earthquake. Near a fault region, earthquakes are very severe, and also near the mid oceanic ridge it is not possible for a person to live. The subduction zones are the region where a denser plate subducts beneath the other, and its impossible for life to exist there.

Thus the most safest place out of all the option is the volcanic islands where numerous people resides. for example, the Hawaiian and Stromboli.

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3 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
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(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

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