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Calculating the density of a sphere (uncertainty included)

  • Thread starter AryRezvani
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  • #1
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Homework Statement



The radius of a solid sphere is measured to be (6.30 ± 0.26) cm, and its mass is measured to be (1.81 ± 0.08) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures. Use the following formula to calculate the uncertainty in the density:
Δρ/ρ = Δm/m + 3Δr/r



Homework Equations



Δρ/ρ = Δm/m + 3Δr/r

D = M/V

The Attempt at a Solution



I'm going to assume you use the radius and calculate the volume of a sphere (4/3pi(r^3), and then convert to m^3.

Use the above information to calculate the density, but how does uncertainty come into play here? Do I plug it into the above equations along with the numbers, or do I do it all at the end with one formula?

I'm a newbie at Physics, really trying to improve myself. Don't laugh guys :(
 

Answers and Replies

  • #2
SammyS
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Homework Statement



The radius of a solid sphere is measured to be (6.30 ± 0.26) cm, and its mass is measured to be (1.81 ± 0.08) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures. Use the following formula to calculate the uncertainty in the density:
Δρ/ρ = Δm/m + 3Δr/r

Homework Equations



Δρ/ρ = Δm/m + 3Δr/r

D = M/V

The Attempt at a Solution



I'm going to assume you use the radius and calculate the volume of a sphere (4/3pi(r^3), and then convert to m^3.

Use the above information to calculate the density, but how does uncertainty come into play here? Do I plug it into the above equations along with the numbers, or do I do it all at the end with one formula?

I'm a newbie at Physics, really trying to improve myself. Don't laugh guys :(
Well, give us some results, & we'll comment.
 
  • #3
67
0
Well, give us some results, & we'll comment.
Alrighty, so I calculated the volume of the sphere and decided to disregard uncertainty until the end of the problem since it has its own formula.

(4/3)*pi*(6.30^3) = 1047.394424 cm^3

Let's just say the results is A

Took A and used dimensional analysis to convert it to m^3

A cm^3 x 1 m^3/1 000 000 cm ^3 = .0010473944

Took the above result and divided it into the mass (1.81), and got 1728 kg/m^3

D = 1728 kg/m^3

--------------------------------------------------------------------------------------------------

A little stumped on the uncertainty formula. What does the P with the curve mean?

And what exactly does it mean by change in P, ect.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,224
947
Alrighty, so I calculated the volume of the sphere and decided to disregard uncertainty until the end of the problem since it has its own formula.

(4/3)*pi*(6.30^3) = 1047.394424 cm^3

Let's just say the results is A

Took A and used dimensional analysis to convert it to m^3

A cm^3 x 1 m^3/1 000 000 cm ^3 = .0010473944

Took the above result and divided it into the mass (1.81), and got 1728 kg/m^3

D = 1728 kg/m^3

--------------------------------------------------------------------------------------------------

A little stumped on the uncertainty formula. What does the P with the curve mean?

And what exactly does it mean by change in P, etc.
Sorry for the tardy reply. :redface:

That's not letter, p. That's Greek letter, ρ (rho) .

ρ in this case is used to represent the density.

So, ρ = m/V. You used the letter, D, for density in your formula, D = M/V.
 

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