Answer:
a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.
b) Average power
P(w)= 1062.07 [w]
P(hp)=1.42 [hp]
c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.
Explanation:
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
Work:

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )
![W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]](https://tex.z-dn.net/?f=W%3D%20F%2Ad%3D%20m%2Ag%2Ad%3D85%2A%209.8%2A12.75%3D10620.75%20%5BJ%5D)
Converting from Joules to Kcals:

Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%

So the student is consuming 12.685 KCals each time he runs up the stairs.
Now,
1 g --> 9 Kcals
1000 g --> 9000KCals
Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

He must run up the stairs 709.5 times, to burn 1 Kg of fat.
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For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)
![Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\](https://tex.z-dn.net/?f=Power%3D%5Cfrac%7BJoules%7D%7BSeconds%7D%20%3D%5Cfrac%7B53103.75%7D%7B50%7D%20%3D1062.075%20%5BW%5D%5C%5C)
![P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]](https://tex.z-dn.net/?f=P%28hp%29%3D%5Cfrac%7BP%28w%29%7D%7B745.7%7D%20%3D%5Cfrac%7B1062.075%7D%7B745.7%7D%20%3D1.42%5Bhp%5D)
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Answer:
P = 1333.33 N
Explanation:
The pressure exerted by the boy on the floor can be calculated by the following equation:

where,
P = Pressure exerted by the boy = ?
F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)
A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)
A = 2(0.06 m x 0.25 m) = 0.03 m²
Therefore,

<u>P = 1333.33 N</u>
Answer:
a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa
, c) F = 20.6 10⁻¹² N
Explanation:
a) The intensity defined as the energy per unit area
S = U / A
Area of a circle is
W = 6.2 mw = 6.2 10-3 W
R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m
A = π R2
A = π (1,080 10⁻⁶)²
A = 3.66 10 -12 m²
S = 6.2 10-3 / 3.66 10-12
S = 1.69 10⁹ W / m²
b) The radiation pressure
P = 1 / c (dU / dt) / A
S = (dU / dt) / A
P = S / c
P = 1.69 10 9 / 3. 108
P = 5.63 Pa
c) the definition of pressure is force over area
P = F / A
F = P A
F = 5.63 3.66 10⁻¹²
F = 20.6 10⁻¹² N
d) for this we use Newton's second law
F = ma
a = F / m