¿Cuál es la magnitud y dirección de la aceleración de los cuerpos en caída libre en el planeta Tierra?
1 answer:
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Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
Answer:
1.75atm
Explanation:
According to Boyle's law, the pressure P of a fixed mass of gas is inversely proportional to it's volume V provided that the temperature remains constant.
This implies the following;
Provided temperature is kept constant.
Given;
From equation (1), we can write;
Since all the units are consistent, there is no need for conversion.
Given
v = 343 m/s
ac = 5g
ac = 5*9.8 m/s^2
ac = 49 m/s^2
where,
v: velocity
ac = centripetal aceleration
Procedure
We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.
Formula
The minimum radius not to exceed the centripetal acceleration is 2401 m.