Respuesta :
The complete question is :
A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .
If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?
Solution :
Given volume of the hot air balloon [tex]$=2100 \ m^3$[/tex]
The balloon fabric weights = 860 N
The weight of the basket with the gear and propane tank = 1300 N
Density of outside air [tex]$= 1.23 \ kg/m^3$[/tex]
∴ Total pay load = Weight of the air displaced - weight of gas inside the balloon
Total pay load = 860 + 1300 + 3400
= 5560 N
Mass = density x volume
Weight [tex]$= \text{mass} \times g$[/tex]
Weight = volume x density [tex]$\times \text{ acceleration due to gravity (g)}$[/tex]
Weight of the displaced air = 2100 x 1.23 x 9.8
= 25313 N
Weight of the gas inside the balloon = density [tex]$\times \text{ acceleration due to gravity (g)}$[/tex] x volume
= density x 9.8 x 2100
= density x 20580 N
Therefore substituting the values, we get
⇒ 25313 - (density x 20580) = 5560
⇒ density [tex]$=\frac{19753}{20580}$[/tex]
[tex]$= 0.96 \ kg/m^3$[/tex]
So the density of the heated gas [tex]$= 0.96 \ kg/m^3$[/tex]
