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zloy xaker [14]
3 years ago
14

A brick lies perilously close to the edge of the flat roof of a building. The roof edge is 50 ft above street level, and the bri

ck has 560.0 J of potential energy with respect to street level. Someone edges the brick off the roof, and it begins to fall. What is the brick's kinetic energy when it is 35 ft above street level?
Chemistry
1 answer:
wel3 years ago
5 0

Hey there!:

The potential energy is U =  m*g*h

Substitute 560.0 J for U = m*g*h so,  560.0 = m*g*h

g = 9.8 m/s²

50 ft in meters :

1 ft = 0.3048 m

50 ft = 50 * 0.3048 => 15.24 m

m = 560.0 / g*h

m = 560.0 / 9.8 * 15.24

m = 560.0 / 149.352

m = 3.74 kg

35 ft in m : 35 * 0.3048 => 10.668 m

The change initial  in potencial energy from point A to B is:

mg ( h1 - h2 ) = 3.74* ( 9.8 )* (15.24 - 10.668 )

=>  36.652 * ( 15.24 - 10.688 ) =

36.652 * 4.552 => 167.57 J

According to the conservation of energy

The change in potencial energy should be equal to the change in kinetic energy , so the , the change in kinetic energy is :

ΔK = 167.57 J


Hope that helps!

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Answer: The correct option is C ( is very hard and burns cleanly).

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The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
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0.0468 g.

Explanation:

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  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

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where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

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8 0
3 years ago
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