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Ivahew [28]
3 years ago
14

If the balloon can barely lift an additional 3500 N of passengers, breakfast, and champagne when the outside air density is 1.23

kg/m3, what is the average density of the heated gases in the envelope
Physics
1 answer:
Bingel [31]3 years ago
8 0

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 $=2100 \ m^3$

The balloon fabric weights = 860 N

The weight of the basket with the gear and propane tank = 1300 N

Density of outside air $= 1.23 \ kg/m^3$

∴  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  $= \text{mass} \times g$

Weight = volume x density $\times \text{ acceleration due to gravity (g)}$

Weight of the displaced air = 2100 x 1.23 x 9.8

                                             = 25313 N

Weight of the gas inside the balloon = density $\times \text{ acceleration due to gravity (g)}$ x volume

                                                             = density x 9.8 x 2100

                                                             = density x 20580 N

Therefore substituting the values, we get

⇒ 25313 - (density x 20580) = 5560

⇒ density $=\frac{19753}{20580}$

                 $= 0.96 \ kg/m^3$

So the density of the heated gas $= 0.96 \ kg/m^3$

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nexus9112 [7]

Answer:

proper time taken by the person is 9.911 × 10⁻⁵ s

Explanation:

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                                                           =  4 × 10⁷ m/s

when the person just passes the street lamp is switched on which is at x =4 km

Lorentz factor = \gamma = \dfrac{1}{\sqrt{1-\dfrac{r^2}{c^2}}}

                                         = \dfrac{1}{\sqrt{1-\dfrac{(4 \times 10^7)^2}{(3 \times 10^8)2}}}

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time taken in your frame of reference,t =\dfrac{D}{v}

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Answer:

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Explanation:

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v_{f}: es la velocidad final = 0 (se detiene)

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a = \frac{v_{f} - v_{0}}{t} = \frac{0 - 22 m/s}{4.0 s} = -5.5 m/s^{2}

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Thus the acceleration of the cabinet=0.956 m/s²

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