Abrick lies perilously close to the edge of the flat roof of a building. the roof edge is 50 ft above street level, and the brick has 370.0 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?

At the roof edge :

we know that 1 ft = 0.305 m

m = mass of the brick = ?

H = height of the edge of roof with respect to street level = 50 ft = 50 (0.305) m = 15.25 m

U = initial potential energy of brick at edge of roof = 370 J

potential energy is given as

U = mg H

inserting the values

370 = m (9.8) (15.25)

m = 2.5 kg

consider the conservation of energy between edge of roof and position of brick at a height of 35 ft :

h = final height above the street level = 35 ft = 35 x 0.305 m = 10.675 m

U = initial potential energy of brick at edge of roof = 370 J

u = final potential energy of brick = mgh = (2.5) (9.8) (10.675) = 261.5375 J

K = kinetic energy at the edge of roof = 0 J      (since it was initially rest)

k = final kinetic energy

using conservation of energy

U + K = u + k

370 + 0 = 261.5375 + k

k = 370 - 261.5375

k = 108.5 J

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

lol for the pointsss