Convex lenses when placed in the air, will cause rays of light (parallel to the central axis) to converge.
Converging lenses, commonly referred to as convex lenses, have thicker centers and narrower upper and lower margins. The edges are outwardly curled. This lens has the ability to concentrate a beam of parallel light rays coming from the outside onto a spot on the opposite side of the lens.
The image created is referred to be a genuine image when it is inverted relative to the object. On a screen, this kind of image can be recorded. When the object is positioned at a point farther than one focal length from the lens, a converging lens creates a true image.
A virtual image is one that cannot be produced on a screen and is formed when the image is upright in relation to the object. When an item is positioned within one focal length of a converging lens, a virtual image is created. It creates an enlarged image of the object on the same side of the lens as the image. It serves as a magnifier.
Learn more about the convex lens here:
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For a standing wave on a string, the wavelength is equal to twice the length of the string:
In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is
And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
according to the facts it should be a numerous amount of wrong answers so ignore dis
