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vitfil [10]
3 years ago
6

Research paper on tsunamis (4-5 paragraphs minimum) 3) Tsunamis how they work….evacuation routes, structures built in oceans to

protect, early detection, etc
Expectations:

A. Must be at least 4 full paragraphs. (probably will need to be more)

B. Must have an introductory paragraph that explains what the problem is, and the general overview of the paper.

C. Must Define Problem, including details and how it relates to Irvine

D. Main Body of Paper (probably couple of paragraphs) This is where you will show what you found out in your research. Make sure to include things like current use/techniques, as well as what technological advancements are being researched to improve your specific topic/issue.

E. Conclusion: Short summary of main points. Your Opinion on what is working and why (or what is not working and why)

F. Citations: Must use at LEAST 3 sources. Must site works at end, in either MLA or APA format, AND must site them where used in the paper (also MLA or APA format).

Make sure all work is yours and original. Copying is cheating. Using information from a source without citing is plagiarism.
Physics
1 answer:
elena55 [62]3 years ago
6 0
A must be at least 4 full paragraphs probably will need more
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Define , gravitational acceleration
Vitek1552 [10]
The simplest answer would be "acceleration due to gravity." 

The exact value of this acceleration changes depending on which planet your on (for example).
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3 years ago
In this lab, you observed how different factors such as velocity, gradient, and ____ , or amount of water in a stream, affect th
egoroff_w [7]

Answer:

volume and erosion

Explanation:

7 0
3 years ago
Read 2 more answers
Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0
3 years ago
Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0
3 years ago
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
4 years ago
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