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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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W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

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(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

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