Lots of reasons. one reason i lie alot (a very bad habit) is im scared of what will happen if i tell the truth. the truth is always better, though.
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)
*****
Answer:
B it decreases
Explanation:
the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy
Answer:
1176.01 °C
Explanation:
Using Ohm's law,
V = IR................. Equation 1
Where V = Voltage, I = current, R = Resistance when the bulb is on
make R the subject of the equation
R = V/I.................. Equation 2
R = 4.3/0.32
R = 13.4375 Ω
Using
R = R'(1+αΔθ)............................. Equation 3
Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature
make Δθ the subject of the equation
Δθ = (R-R')/αR'.................. Equation 4
Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹
Substitute into equation 4
Δθ = (13.4375-1.6)/(1.6×0.0064)
Δθ = 11.8375/0.01024
Δθ = 1156.01 °C
But,
Δθ = T₂-T₁
T₂ = T₁+Δθ
Where T₂ and T₁ = Final and initial temperature respectively.
T₂ = 20+1156.01
T₂ = 1176.01 °C
Answer: It should be A or the very left red circle that you can click on
Explanation: Because when the wind is moving downward and the earth is spinning the spot the wind ends up will never be directly down from where it was to begin with
