No dia 19 de outubro de 1901, Santos Dumont, utilizando o seu dirigível número 6, circundou a Torre Eiffel, ganhando, assim, o Prêmio Deutsch, oferecido pelo milionário Henry Deutsch. Sabe-se que um dirigível utiliza um grande volume de gás menos denso que o ar, para que seu empuxo fique maior que o seu próprio peso, fazendo assim com que ele suba.
Sendo ρar a massa específica do ar, ρd a massa específica do dirigível, g a aceleração da gravidade, qual à aceleração de um dirigível ao subir em função de ρar, ρd e g, sendo que as únicas forças agindo nele são o peso e o empuxo? Desconsidere qualquer tipo de forças resistivas.
- A)
large a = g(1- {par over pd})
- B)
large a = g({par over pd} - 1)
- C)
large a = g({pd over par} - 1)
- D)
large u = {({par over pd} - 1) over g}
Resposta:
Let's break down the question step by step. We are given that the airship uses a large volume of a less dense gas than air, a characteristic that allows it to rise. This is because the upward buoyant force (empuxo) on the airship is greater than its weight.
Now, let's analyze the situation using physics. We can use Archimedes' Principle, which states that the buoyant force (empuxo) on an object is equal to the weight of the fluid (in this case, air) displaced by the object. Since the airship is rising, we know that the buoyant force is greater than the weight of the airship.
Let's denote the mass of the air as ρar and the mass of the airship as ρd. We know that ρar is less than ρd, which is why the airship rises. The acceleration of the airship due to gravity is g, and the acceleration of the airship as it rises is a.
Using Newton's second law, we can write the equation:
$a = g left(1 - frac{rho_a r}{rho_d}right)$
This equation shows that the acceleration of the airship is directly proportional to the difference between the densities of the air and the airship.
Now, let's analyze the options:
- A) $a = g left(1 - frac{rho_a r}{rho_d}right)$ (correct answer)
- B) $a = g left(frac{rho_a r}{rho_d} - 1right)$
- C) $a = g left(frac{rho_d}{rho_a r} - 1right)$
- D) $u = frac{left(frac{rho_a r}{rho_d} - 1right)}{g}$
The correct answer is B) $a = g left(frac{rho_a r}{rho_d} - 1right)$. This equation shows that the acceleration of the airship is directly proportional to the difference between the densities of the airship and the air, which is consistent with our previous analysis.
Therefore, the correct answer is option B) $a = g left(frac{rho_a r}{rho_d} - 1right)$. This equation correctly describes the acceleration of the airship as it rises, taking into account the difference in densities between the air and the airship.
Deixe um comentário