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4 years ago

Answer the following question about "The Deadliest Tsunami in History."

Physics

2 answers:

Anna007 [38]4 years ago

7 0

The correct answer is - a. was a sign of danger.

Once the people saw that the ocean waters are receding and were living vast space without water behind them, they knew that something big and very dangerous will happen. And in fact it did. The water that was sucked in in the place were there was a crack on the ocean floor, got shot back under big pressure and it had very big speed, as well as having waves that were destroying anything on their way.

NeTakaya4 years ago

3 0

Answer:

Option (A)

Explanation:

Tsunamis are the series of large waves that are generated in the large water bodies such as seas and oceans, due to an earthquake, underwater explosion or landslide. These waves can rise up to a height of about hundreds of meters.

A receding ocean is a sign of danger that signifies that a tsunami is approaching. During this time the water level drops down unusually along the coastline. It is a kind of warning that is given to the people living in the coastal areas to evacuate it as soon as possible in order to save their lives.

Thus, the correct answer is option (A).

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Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangmen

murzikaleks [220]

Answer:

a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d) W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

W = ΔU = U_∞ -U

the potential energy for point charges is

U =k $\sum \frac{q_i q_j}{r_{ij} }$

in this case we only have two particles

U = k $\frac{q_1q_2}{r_{12} }$

the distance is

r₁₂ = $\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }$

r₁₂ = $\sqrt{ 0 + ( 2-0)^2}$ Ra 0 + (2-0)

r₁₂ = √2= 1.4142 m

we substitute

W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

W = $\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}$ 9 109 1.6 10-19 1.6 10-19 / 1.4142

W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

U = k $( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )$

in this case all charges are electrons

q₁ = q₂ = q₃ = q

W = U = k q² $( \frac{1}{r_{13} } + \frac{1}{r_{23} } )$

the distances are

r₁₃ = $\sqrt{(3-0)^2 + 0}$ RA (3.00 -0) 2 + 0

r₁₃ = 3

r₂₃ = $\sqrt{ 3^2 + 2^2}$ Ra (3 0) 2 + (2 0) 2

r₂₃ = √13

r₂₃ = 3.606 m

let's look for the job

W = U

let's calculate

W = ${9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )$

W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

W = U = k $( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )$

all charges are equal q₁ = q₂ = q₃ = q₄ = q

W = k q² $(\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )$

let's look for the distances

r₁₄ = $\sqrt{3^2 +4^2}$

r₁₄ = 5 m

r₂₄ = $\sqrt{3^2 + ( 4-2)^2}$

r₂₄ = √13 = 3.606 m

r₃₄ = $\sqrt{(3-3)^2 + (4-0)^2}$

r₃₄ = 4 m

we calculate

W = 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )$

W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

W = U = k $( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )$

in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value

W = - k q² $( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )$

we look for distances

r₁₅ = $\sqrt{ 1^2 +1^2}$ Ra (1-0) 2 + (1-0) 2

r₁₅ = √ 2 = 1.4142 m

r₂₅ = $\sqrt{ (2-1)^2 +1^2}$

r₂₅ = √2 = 1.4142 m

r₃₅ = $\sqrt{ ( 3-1)^2 +1^2}$

r₃₅ = √5 = 2.236 m

r₄₅ = $\sqrt{ (3-1)^2 + (4-1)^2}$

r₄₅ = √13 = 3.606 m

we calculate

W = - 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )$

W = - 4.93 10⁻²⁸ J

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3 years ago

What are tree things that apply to relative dating

grigory [225]

Rocks, earth aging, fossils

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3 years ago

A duck floating on a lake oscillates up and down

slega [8]

The answer is (2) 0.50 Hz. The frequency (f) of oscillation is the number of oscillations (n) per time (t) in seconds: f = n/t. A duck floating on a lake oscillates up and down 5.0 times (n = 5.0) during a 10.-second interval (t = 10.0 s). So, the frequency of duck's oscillations is: f = 5.0/10.0 s = 0.50 1/s = 0.50 Hz.Hope I helped! :) Cheers!

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3 years ago

Each roller under a conveyor belt has a radius of 0.5 meters. the rollers turn at a rate of 30 revolutions per minute. what is t

svp [43]

Distance in a minute=0.5 times 30=15 meters
distance in a second=15 divided by 60=0.25 meters per second
hope it helps

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3 years ago

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Jimena is stopped at a red light. When the light turns green, she

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