Astudent evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of steps in order to lose one pound (0.45 kg) of fat. metabolizing 1.00 kg of fat can release 3.77 x 10' 1 of chemical energy and the body can convert about 22.0% of this into mechanical energy (the rest goes into internal ) how much mechanical energy (in 3) can the body produce from 0.450 kg of ) how many trips up the flight of steps are required for the 68.0 kg student to lose 0.450 kg of fat? ignore the relatively small amount of energy required to return down the .

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).