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kodGreya [7K]
1 year ago
6

Read the following questions and answer them using complete sentences. Be sure to fully explain your answers.

Physics
1 answer:
astra-53 [7]1 year ago
6 0

Wave power can be regarded as a reliable source of energy because the ocean currents are always moving.

<h3>What can be the challenges of wave power?</h3>

Wave power is a device that can be used to convert the mechanical energy of the ocean waves into electrical energy based on the principle of conservation of energy.

The major challenges that face the use of wave power in electricity generation is the unreliability of the waves which leads to uncertainty in the quantity of power generated Also, the wave direction and direction of ocean currents all limit the amount of power generated by this method. However, in spite of challenges, it can be regarded as a reliable source of energy because the ocean currents are always moving.

Learn more about wave power:brainly.com/question/1362067

#SPJ1

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Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
2 years ago
Pour chacune des réactions chimiques
Doss [256]

Answer: Me non espanol

Explanation:

4 0
2 years ago
I need help ASAP I need to get this right plz plz plz!!!!!
Rina8888 [55]

<em><u>The</u></em><em><u> </u></em><em><u>atomic</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>consists</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>protons</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u>.</u></em>

<em><u>Additional</u></em><em><u> </u></em><em><u>information</u></em><em><u>:</u></em>

<em><u>Protons</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>positive</u></em><em><u>ly</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particl</u></em><em><u>e</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>negative</u></em><em><u>ly</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u>.</u></em>

<em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>:</u></em><em><u>)</u></em>

3 0
3 years ago
Read 2 more answers
A crate falls from an airplane flying horizontally at an altitude of 2,000 m.
Hatshy [7]

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

7 0
3 years ago
I NEED HELP ITS AN ECON QUESTION THE PICTURE IS ABOVE IF YOUR CORRECT I WILL GIVE YOU THE EXTRA POINTS APPRECIATE THE HELP
tester [92]

Answer:

choose to spend limited resources to meet their needs

3 0
3 years ago
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