Respuesta:
Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.
Explicación:
La masa total que el avión es capaz de levantar es:
Para el aluminio:
y
![V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]](https://tex.z-dn.net/?f=V_%7Bfuselage%7D%3D%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D)
donde:
- L es longitud
- D es diámetro
- e es grosor
![m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BAl%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Al%7D)
Para el acero (mismo procedimiento):
![m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BSteel%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Steel)
Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.
También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.
Greetings!
The result is:
The new volume is: 
Rationale:
Because the temperature remains constant, we can apply Boyle's Law to solve this issue.
Boyle's Law stipulates that:
Where,
P is the gas's pressure.
V is the gas's volume.
According to the information provided:
Let's put the values into the equation:
Consequently, the new volume is:
Wishing you a lovely day!
Answer:
Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.
Explanation:
Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
