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masha68 [24]
3 years ago
6

Research the potential for a large earthquake off the coast of Oregon. What are two of the biggest concerns with a large earthqu

ake in this area?
Physics
1 answer:
Sphinxa [80]3 years ago
6 0

There is a 33% chance of a large earthquake off the coast of Oregon.

The two main concerns of a large earthquake are Gas fires and Tsunami .

Explanation:

Everytime a large earthquake hits in any other part of the world, We find a small earthquake hitting off the coast of Oregon.

So scientists feel that as United States is prone to Earthquakes, the people have to be prepared for such eventualities.

The main concern for this large earthquake is, that it might lead to gas leakages at homes that can start a gas fire, which might spread. So people have to be careful and check  on any gas leakages.

Such large earthquakes can also lead to tsunami waves. So people have to take precautions and evacuate, if they are near the coastline.

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Please help me!! Time = 98 seconds speed=7.5 meters/second what is the distsnce in meters
solmaris [256]

Answer:

<h2>735 m</h2>

Explanation:

The distance covered by an object given it's speed and time taken can be found by using the formula

distance = speed × time

From the question we have

distance = 7.5 × 98 = 735 m

We have the final answer as

<h3>735 m </h3>

Hope this helps you

3 0
3 years ago
George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. Wh
sergeinik [125]
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4 years ago
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A ball is dropped from a height of 1.30 m. How long does it take to hit the ground? WebAssign will check your answer for the cor
Citrus2011 [14]

Answer:

a) t= 0.515 s

b) vf= 5.047 m/s

Explanation:

Because ball move with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*g*t² Formula (1)

vf= v₀+gt Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Known data

v₀=0

d= 1.30 m

g= 9.8 m/s²

Problem development

a) We appy the formula 1 to calculate the time (t):

d= v₀t+ (1/2)*g*t²

1.30= 0+ (1/2)* (9.8)*t²

1.30=4.9*t²

t²=1.30/4.9

t=\sqrt{\frac{1.3}{4.9} }

t= 0.515 s

b) We appy the formula 2 to calculate the final speed (vf):

vf= v₀+gt

vf= 0+(9.8)*(0.515)

vf= 5.047 m/s

8 0
3 years ago
Question: As the moon spins on its axis we see
butalik [34]

Answer:

<h2><em><u>A) Different parts of the moon throughout the month</u></em></h2>

Explanation:

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3 years ago
Waves transport
Brilliant_brown [7]

Answer:

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Explanation:

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