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allochka39001 [22]
3 years ago
9

Which does not provide the energy to spin a turbine?

Physics
2 answers:
Sophie [7]3 years ago
6 0
It depends on what kind of the turbine to use it could be electromagnet or water in reservoir
Makovka662 [10]3 years ago
3 0

It's B, Electromagnets.

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A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas
goldfiish [28.3K]
    <span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
3 0
3 years ago
A TV satellite dish is designed to receive radio waves of wavelength
Verizon [17]

Answer:

4.7 GHz

Explanation:

Applying,

v = λf................. Equation 1

Where v = velocity of the radio wave, λ = wavelength, f = frequency

make f the subject of the equation

f = v/λ.............. Equation 2

Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s

From the question,

Given: λ = 0.0644 meters

Constant: v = <em>3.00 x 10⁸ m/s</em>

Substitute these values into equation 2

f = (3.00 x 10⁸)/0.0644

f = 4.66×10⁹ Hz

f = 4.7 GHz

3 0
3 years ago
In 1965, a group of students wore black arm bands to school in protest of American policies in Vietnam. Administrators banned th
Ede4ka [16]

Answer:

protected under students first amendment rights

Explanation:

did the studyisland :)

4 0
3 years ago
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
3 years ago
You hear the sound from a tuning fork vibrating at 483 Hz. You then hear the sound from a second tuning fork vibrating at 512 Hz
AleksAgata [21]

Answer:

tgggggg

Explanation:

6 0
3 years ago
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