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Murljashka [212]
3 years ago
8

What is the key difference between mechanical and electromagnetic waves

Physics
1 answer:
Alecsey [184]3 years ago
7 0
Mechanical waves require a medium to travel, but electromagnetic waves do not.
You might be interested in
The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig
julia-pushkina [17]

Answer: They are all true

a. The tension in the rope is everywhere the same.

b. The magnitudes of the forces exerted on the two objects by the rope are the same.

c. The forces exerted on the two objects by the rope must be in opposite directions.

d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)  

- Cutiepatutie ☺❀❤

5 0
3 years ago
An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor
MrMuchimi

The electric flux through the hole is 56.45\ webber .

  • Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
  • Mathematically it is given by \phi_E=E.A \ Nm^2/C where E is the electric field and A is the area.
  • Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by  Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

The whole surface of hollow sphere = 4\pi r^2

                                                            = 4\times 3.14\times  (10 \times  10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2

Area of the hole ( both side ) = 2\times \pi  r^2

                                               = 2\times 3.14 \times  (10^-^3)^2\\= 6.28 \times 10^-^6 m^2

According to Gauss's theorem, the flow from a particular charge in the center is given by

 \phi=  \frac{10\times10^-^6}{8.85\times 10^-^1^2}\\\\\phi=1.13\times10^6

This flux flows through the surface of the sphere, so the flux  per unit area which is given by

= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \  weber / m^2

Flux through area of hole is given by :

=  8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber

Learn about more electric flux here :

brainly.com/question/26289097

#SPJ4

8 0
1 year ago
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of
Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

\eta = 1-\frac{T_L}{T_H}

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

4 0
2 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
2 years ago
if researchers hit a sheet of metal with a yellow light and nothing happens, what color should the try next?​
Nataliya [291]

The photoelectric emission is possible if the wavelength of the incident light is less than that of yellow light

3 0
3 years ago
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