How Big Does a Helium Balloon Need to Be to Reduce a Person's Weight by 21%?

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    Buoyancy Volume
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Homework Help Overview

The discussion revolves around calculating the volume of a helium balloon needed to reduce a person's weight by 21%. The original poster presents a scenario involving a friend with a mass of 79 kg and seeks to determine the appropriate balloon size in cubic meters.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various calculations related to buoyancy and density, questioning the assumptions made regarding air and helium densities. Different volume estimates are presented, with some participants expressing uncertainty about the accuracy of their results.

Discussion Status

The conversation is ongoing, with participants providing calculations and corrections to each other's approaches. Some guidance has been offered regarding the necessary buoyancy and density values, but there is no clear consensus on the correct volume required for the balloon.

Contextual Notes

Participants note the importance of considering conditions such as altitude and temperature, which may affect density values. There is also mention of a homework program that may have specific requirements for the answers submitted.

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Homework Statement


"friend with a mass of 79 kg ponders the idea of attaching a helium-filled balloon to himself to effectively reduce his weight by 21% when he climbs. He wonders what the approximate size of such a balloon would be"
solution is expressed as cubic meters
apology if this entry is miscategorized

Homework Equations

volume =mass/density

The Attempt at a Solution

volume= change in mass/ difference in density=16.59/(1.2-1.78)=16.27m^3
this amount is inaccurate, as is 13.27 m^3
 
Last edited:
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You have the right idea, you just need to be a bit more careful about working it out.

You want a bouyancy of 21% * 79kg = 16.59kg
bouyancy = volume * difference in density

Density of air = 1.27 kg/m^3 helium = 0.18 kg/m^2
 
presuming:
16.59 kg/1.09 kg/m^3=15.22 m^3 (volume)
neither 15 nor 15.22 m^3 is the correct solution- according to my online homework program
 
Last edited:
Unless it gives you a value for the altitude or temperature, then assuming STP
Density 1.2754 - g/l (air) 0.1786 g/l (he) = 1.0968 g/l (or kg/m^3)
You need to lift 79*0.21 kg / 1.0968 kg/m^3 = 15.1 m^3
 
Last edited:
these are the incorrect answers submitted:
13
13.2
13.27
13.28
16.95
16.23
16.26
1.38
16
18.08
18
15.22
15...14.9...15.1
 
Last edited:
exact question:
"A mountain-climber friend with a mass of 79 kg ponders the idea of attaching a helium-filled balloon to himself to effectively reduce his weight by 21% when he climbs. He wonders what the approximate size of such a balloon would be. Hearing of your physics skills, he asks you. What answer can you come up with, showing your calculations?"
-
solution is in m^3

-thank you for the explanation, i have no more available attempts...should find out the correct answer tomorrow, will update
 
Last edited:
according to Paul G. Hewitt, via webassign.net;
the solution is:
volume = 79 kg*.21/1.2 kg/m^3= 13.825 m^3
no consideration of helium density and air density is considered 1.2
 

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