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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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Which career professionals are part of the Architecture and Construction career cluster? Check all that apply.
oksian1 [2.3K]

Answer:

A,B,D,E,F

Explanation:

I took the test for yall.

3 0
3 years ago
When working with a powered instrument, the swirling effect produced by the water stream within the confined space of a periodon
Lina20 [59]

Answer:

Acoustic microstreaming

Explanation:

Acoustic microstreaming is the swirling effect produced by water stream confined in a spaced of a periodontal pocket.

  • It is the movement of water in a particular direction as a result of mechanical pressure within the fluid body.
  • They are often used in dental procedures to remove particulates from the teeth.
  • It mostly relies on the properties of sound waves to achieve this goal
3 0
3 years ago
A simple pendulum of length of 1.37 m and mass of 6.66 kg is given an initial speed of 2.85 m/s at its equilibrium position. Det
yawa3891 [41]

Answer:

2.35 s

Explanation:

The period of a simple pendulum is expressed as;

                                T = 2π\sqrt{\frac{L}{g} }

Where

T is the period in seconds

L is the length in metres

g is acceleration due to gravity

                                 T = 2π\sqrt{\frac{1.37}{9.8}}

                                 T = 2.349 s

                                 T = 2.35 s

8 0
3 years ago
Can anyone help me with questions a and c​
Ganezh [65]

Answer:

a)

n=sin i/sin r

n= -0.305/-0.428

n=0.713

b)

sin c=1/n

sin c=1/0.713

sin c= 1.403

c=sin⁻¹(1.403)

c= 40.842°

Explanation:

i hope it will be helpful

plzzz mark as brainliest

5 0
3 years ago
The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con
ira [324]

The Period of the resulting shm will be T=39.7

<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

ω=k/m

ω=1200/3

ω=400

<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the  torque value </u>

τ = -dmgsinΘ    where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

<u>Rotational to linear form</u>

a = αr  

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

6 0
3 years ago
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