Answer:
770m/s
Explanation:
caculation using one of the newton law of motion
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.
********************
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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The spontaneous emission of radiations from an unstable nuclei is known as natural radioactivity. on the other hand, The process of emission of radiations from naturally occurring isotopes when they are bombarded with sub-atomic particles or high levels of X-rays or gamma rays called artificial radioactivity.
Recall this kinematic equation:
a = 
This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).
a is the acceleration.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 25 m/s.
Δt = 10 s
Substitute the terms in the equation with the given values and solve for a:
a = 
<h3>a = 2.5 m/s²</h3>
Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.


W is the work done by the friction.







v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.