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Increase your percent confidence to provide an increased width.
56
This needs more information. Without some other factor, like a change in area, the width doesn't have to increase at all.
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The perimeter decreases by 4. Decreasing the length by 4 decreases the perimeter by 8. Increasing the width by 2 increases the perimeter by 4. -8+4=-4 For example: Area of a rectangle could be: 8 x 6 (8 being the length and 6 being the width). The perimeter is 8+8+6+6=28 If the length is decreased by 4 then it becomes 4 If the width is increased by 2 then it becomes 8. The perimeter becomes 4+4+8+8=24
Fringe width is a term used for the width of the first maxima in diffraction. It can also be found by calculation by dividing the wavelength of the light input multiplied by the distance between the slits and the screen by the space between the slits.
if the width of one slit is increased relative to the other the slit separation must decrease and since slit sep is inversely proportional to fringeseparationthe fringes become closer together.
Silicon, Germainium, Tin, Lead, Ununquadium.
it is costruacted by an arrangement consisting of a large number of parallel slits of a same width and separated by equal opaque spaces .
When point sources are moved further apart, there is a greater amount of interference produced, as evidenced by a larger number of nodes. When the amount of interference increases, the width of any given antinode decreases.
it is costruacted by an arrangement consisting of a large number of parallel slits of a same width and separated by equal opaque spaces .
increased rms value
Increase your percent confidence to provide an increased width.
decreases
depletion layer decreases
Both the width and length have been increased by 4 feet and so 9*14 = 126 square feet
32m2 = 2m x 16m(width)Length(a) x width (16m) = 288m2288m2 / 16m = 18m