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Vanyuwa [196]
3 years ago
13

Large waves produced by earthquakes

Physics
2 answers:
galina1969 [7]3 years ago
8 0

Answer:

A

Explanation:

If the earthquake occus in the ocean, it can push up powergul waves.

Diano4ka-milaya [45]3 years ago
4 0

Large waves produced by earthquakes is known as Tsunami.

<h3><u>Explanation:</u></h3>

In an ocean or lakes, sometimes a series of large waves called tsunamis are formed which is caused to the displacement of large amount of water. The reason for these tsunamis are earthquakes and other factors like volcanic eruptions and other underwater triggers.

The earthquakes trigger the waves to rise to a large height and rush into inlands with intensity, which often causes large scale destruction. The earthquakes causes displacement in the plates of the earth which makes a large scale vibration which results in tsunamis.  

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3 years ago
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A velocity-time graph is shown below: (2 points)
BabaBlast [244]

Answer:

2m/s²

Explanation:

  • V=Vstart+at
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part A of graph

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part B of graph

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the average between the two is both answers added divided by the number of answers

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5 0
3 years ago
A 2 kg ball is thrown down with 50J of energy from a height of 10m, what is its velocity before it strikes the ground (neglect a
pochemuha

Answer:

V=14

Explanation:

PE=KE

mgh=1/2mv^2

2(9.8)10=1/2(2)v^2

(radical) 196= (radical)v

V=14

7 0
3 years ago
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
Ad libitum [116K]

Answer:

The thickness of the oil slick is 1.95\times10^{-7}\ m

Explanation:

Given that,

Index of refraction = 1.28

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Order m = 1

We need to calculate the thickness of oil slick

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2nt= m\lambda

Where, n = Index of refraction

t = thickness

\lambda = wavelength

Put the value into the formula

2\times1.28 \times t=1\times\times500\times10^{-9}

t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }

t=1.95\times10^{-7}\ m

Hence, The thickness of the oil slick is 1.95\times10^{-7}\ m

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3 years ago
At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

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v is 101m/s

substituing the parameters into the equation, to get the radius

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Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0
4 years ago
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