Question

A lighter-than-air spherical balloon and its load of passengers and ballast are floating stationary above the...

A lighter-than-air spherical balloon and its load of passengers and ballast are floating stationary above the earth. Ballast is weight (of negligible volume) that can be dropped overboard to make the balloon rise. The radius of this balloon is 7.62 m. Assuming a constant value of 1.29 kg/m3 for the density of air, determine how much weight must be dropped overboard to make the balloon rise 129 m in 19.0 s.

Homework Answers

Answer #1

This isn't going to be exactly right, because we don't know what's in the balloon to make it lighter than air. However neglecting that fact we do the following.

Step One

=======

Find the upward acceleration.

d = 129 m

a = ???

vi = 0

t = 19 sec

d = vi * t + 1/2 at^2

129 = 1/2 a * 19^2

a = 0.715 m/s^2

Step Two

=======

Find the Volume of the balloon and from that the mass.

V = 4/3 * pi * r^3

r = 7.62 m

V= 1853.3 m^3.

density_of_air = m/V

1.29 = m/1853.3

m = 2391 kg

Step Three

=========

The force keeping this in equilibrium =

F = m* g

F = 2391 * 9.81 = 23453 N

Step Four

=======

Find the mass needed to cause the upward acceleration.

23453 = (2391 - x) * (9.81 + 0.715)

23453 = (2391 - x) * 10.525 Divide both sides by 10.525

2229 = 2391 - x

x = 162.69 kg

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