Answer:
∴ [T]=[WF−1V−1]
Hope this answer is right!!
This applies to nuclear reactions, specifically nuclear fission.
This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.
In some unusual applications of unusual components, I can think of unusual electric circuits where a switch may be connected in parallel with a device in order to control it.
But I'm sure this is not what's intended in a question on the high-school level.
Until you get in a situation with tricky applications in a tricky circuit, your switches will always be connect <em>in series</em> with the devices they control.
Acceleration is the rate of change of velocity
To define acceleration, We need to know more about motion.
Motion: This can be defined as the change in position of a body from one point to another. When an object accelerates, it undergoes motion.
<u>Definition</u>
Acceleration can be defined as the rate of change of velocity. The S.I unit of acceleration is meter-per-squared seconds. (m/s²)
The formula of acceleration is
- a = (v-u)/t................. Equation 1
⇒ Where:
- a = acceleration
- u = initial velocity
- v = final velocity
- t = time
Hence, Acceleration is the rate of change of velocity
Learn more about acceleration here: brainly.com/question/605631
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)
*****