Which half reaction occurs at the cathode in this cell? 2003--O2 + 2002 + 4e-
+ 2002 + 4e--> 2003-
2H2 +2003-2H2O + 2002 +4e-
O 2H2O + 2002 +40-—— 2H2 +2003-

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

the answer would be c

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.