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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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An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you
Ilya [14]

Answer:

P = 11666.6 W

Explanation:

Given that,

Work done by the motor, W = 3500 kJ

Time, t = 5 min = 300 s

We need to find the power developed by the motor. Power developed is given by :

P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W

So, the required power is 11666.6 W.

4 0
3 years ago
A woman drives 6 miles, accelerating uniformly from rest to 70 mph. How long does it take for her to reach 70 mph?
Morgarella [4.7K]
Using the two kinematic equations that can be used for this problem are:
Vf = Vi + at and d=Vit +(1/2)*at^2
Since Vi (initial velocity) = 0 

The equations can further be simplified where a is the acceleration, t is the time, Vf is the final velocity which is 70 miles per hour and d is 6 miles

Vf = at
70 = at
a = 70/t---equation 1

d=(1/2)*a*(t^2)
6 = (1/2)*a*(t^2) ---equation 2

Substituting equation 1 to equation 2.
6= (1/2)*(70/t)*(t^2)
6= 35t
t= 0.17142 hours or 10.28571 mins or 617.14 sec



6 0
3 years ago
A (nonconstant) harmonic function takes its maximum value and its minimum value
AURORKA [14]

Answer:

Explanation:

Consider that F (any function) <0 .

u(x,y) is a coontinuous function in the closed interval or region R.

Let us consider a point (p,q) that is inside the region and it is a maximum point.

Then it should be must

uxx (p,q) <0 where uxx means double differentiation

and uy(p,q) >0

Since ux(p,q) = 0 = uy(p,q) where ux and uy means single differentiation with respect to x and y respectively.

Say, Maximum limits of the region is T

therefore q<T

then uy (p,q) = 0 if q<T

if q = T then

point (p,q) = (p,T) will be on the boundary of R then we claim that

uy(p,q) >0

Similarly for the minimum also it will work

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3 years ago
What is Archimedes principles ???&amp;<br><br><br><br>yaha avo sab ​
vova2212 [387]

Answer:

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7 0
3 years ago
Read 2 more answers
A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part
Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

(I_p)s = \frac{\pi(0.55)4}2

      = 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

T_{max} = T_{allow} \frac{I_p}{r}

         =(14 \times 10^3) \times (\frac{0.14374}{0.55})

         = 3658.836 lb.in

         = \frac{3658.836}{12} lb.ft

        = 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

T_{max} = T_{allow}( \frac{Ip}{r})

          = (14 \times10^3) \times ( \frac{0.13101}{0.55})

          = 3334.8 lb.in

          = (\frac{3334.8}{12} ) lb.ft

          = 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

P_{max} = 2 \pi f_T

          = 2\pi(2.1) \times 304.9

          = 4023.061 lb. ft/s

          = (\frac{4023.061}{550}) hp

          = 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

P _{max,t} = 2\pi f_T

            = 2\pi(2.1) \times 277.9

            = 3666.804 lb.ft /s

            = (\frac{3666.804}{550})hp

            = 6.667 hp

7 0
4 years ago
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