Respuesta :
Answer:
a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]
Explanation:
Let's solve this problem in parts, with the initial data the camera rises slowly, so we can assume that at constant speed, we apply the equilibrium condition
B - W = 0
The thrust is given by Archimedes' law
B = rho_aire g V_body
The volume of the body can be found from the ideal gas ratio
P V = n R T
let's use the subscript "o" for the initial concisions
V₀ = (nR) T₀/P₀ 1
we substitute
10 ρ_air g (nR) T₀ /P₀ = W
when the storm approaches the pressure decreases, P
If we use the ideal gas equation
V = (nR) T₀ / P
we combine this equation with equation 1
V = V₀P₀ / P
if we write the pressure
P = P₀ -ΔP
we substitute
V = [tex]\frac{V_oP_o}{P_o - \Delta P} =V_o \ ( 1 - \frac{\Delta P}{P_o} )^{-1}[/tex]
we expand serie and eliminate higher order terms
V = V₀ ([tex]1+ \frac{\Delta P}{Po}[/tex])
with this expression we can write the thrust
B = B₀ + ΔB
Newton's second law for the new conditions is
B - W = m a
(B₀ + ΔB) - W = ma
ΔB = m a
a = ΔB / m
a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]
This is the initial acceleration of the camera
