-- The car starts from rest, and goes 8 m/s faster every second.
-- After 30 seconds, it's going (30 x 8) = 240 m/s.
-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s
-- Distance covered in 30 sec at an average speed of 120 m/s
= <span> 3,600 meters .</span>
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The formula that has all of this in it is the formula for
distance covered when accelerating from rest:
Distance = (1/2) · (acceleration) · (time)²
= (1/2) · (8 m/s²) · (30 sec)²
= (4 m/s²) · (900 sec²)
= 3600 meters.
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When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.
When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.
How does he do that ?
By accelerating at 8 m/s². That's about 0.82 G !
He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !
He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
If you mean like electromagnetic waves then, Mico waves, UV rays, and infrared waves
To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:
Where,
= Wavelength
d = Width of the slit
= Angular resolution
Through the arc length we can find the radius, which would be given according to the length and angle previously described.
The radius of the beam on the moon is
Relacing
Replacing with our values we have that,
Therefore the diameter of the beam on the moon is
Hence, the diameter of the beam when it reaches the moon is 7361.82m
Answer:
Explanation:
If a perfect vacuum existed in any volume, then no sound would be able to propagate through it, because a sound wave is a pressure wave, and there would be identically zero pressure. Of course, we could get into speculations about “dark energy” or “vacuum energy” supporting pressure waves, but let’s not go there.
