In our current view of the Universe there are 2 possible futures. One possiblity is that the Universe will come to an end in the opposite of a Big Bang called The Big Crunch. The other possibility is that we live in an Eternal Universe that will never come to end.
i hope this helps :)
Answer:
A
Explanation:
The line(A) goes throughout the entire picture. So therefore choice A would be it's length.
There are two<span> main types of </span>wave<span> interference: constructive interference and destructive interference. Constructive interference </span>happens<span> when the amplitude of the combined </span>waves<span> is larger than the amplitudes of the single </span>waves<span>. This can occur when the </span>crests of two<span> transverse </span><span>waves overlap.
Hope this helps!!! :D
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Answer: maximum height= 40.8m
Explanation: shown in the attachment.
Goodluck
A) 750 m
First of all, let's find the wavelength of the microwave. We have
is the frequency
is the speed of light
So the wavelength of the beam is

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find

and so, the diameter is

B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

So the area is

And since the power is

The average intensity is
