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Jobisdone [24]
3 years ago
6

A ball of mass 8 kg falls from rest from a height of 100 m. Neglecting air resistance,calculate its kinetic energy after falling

a distance of 30 m.
Physics
2 answers:
astra-53 [7]3 years ago
4 0
As an object falls from rest, its gravitational energy is converted to kinetic energy

G.P.E = K.E = mgh

K.E = (80 Kg)(9.8 m/s²)(30 m)

K.E. = 23,520 J
xxMikexx [17]3 years ago
3 0
At h=100m,
ME = KE + PE
ME = 1/2mv^2 + mgh
ME = 1/2m(0)^2 + mgh     Because the ball has no speed at first
ME = mgh = 8 x 9.8 x 100 = 7840 J

At h=70m   (100m-30m)
ME = KE + PE
7840 J = KE + mgh
KE = 7840 J - mgh = 7840 - 8 x 9.8 x 70
KE = 7840 - 5488
KE = 2352 J

Hope this helps

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Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

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