A100 kg individual consumes 1200 kcal of food energy a day. calculate
(a) the altitude change, in m, if the food energy content was converted entirely into lifting the indi-
vidual under normal earth gravity.
(b) the velocity, in m/s, if the food energy content was converted entirely into accelerating the indi-
vidual from rest.
(c) the final temperature, in ◦c, of a 100 kg mass of liquid water initially at the normal human body
temperature and heated with the energy content of the food. you can use a liquid water specific
heat of 4.1 kj/kg k.

(a) 5142.86 m

(b) 317.5 m/s

(c) 49.3 degree C

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 =  1/2 x 100 x v^2

v = 317.5 m/s

(c) The temperature of normal human body, T1 = 37 degree C

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s