Determining the Density of a sphere.

In summary, the radius of a sphere is 6.90 cm and its mass is 1.75 kg. The sphere is solid and has a density of 1271.8 kg/m^3.
  • #1
teamsoulglo
2
0

Homework Statement



The radius of a sphere is measured to be (6.90 0.25 ) cm, and its mass is measured to be (1.75 0.05) kg. The sphere is solid. Determine its density in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures.)

Homework Equations



Volume = (4/3) (pi)(radius)^3

= (4/3) (3.1416)(6.9/100)^3 = 1.376 X 10^-3 m^3

Density = mass / vol = 1.75 / 1.376 X 10^-3 = 1271.8 kg / m^3

= 1.27 X 10^3 kg /m^3


The Attempt at a Solution



% error in radius = (0.25/6.9)X100 = 3.6%
therefore % error in volume = 3 X 3.6% = 10.8%
% error in mass = (0.05/1.75) X 100 = 2.86%

So % error in the density = 10.8 + 2.86 = 13.66%

Therefore undertainty in density = 13.66% of 127 X 10^3 kg/m^3

= 173 kg/m^3

So density = (1.27±0.173) X 10^3 kg/m^3

According to webassign, my answer is incorrect. I can't figure out what I did wrong. Any help would be appreciated.
 
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  • #2
My first time using this board, and no response for over 4 hours. Yet others who post after me are getting help asap. Thanks for the help guys!
 
  • #3
Welcome to the forums teamsoulglo,

Patients is a virtue, and besides, I'd be impressed if on average questions were answered quicker on any other boards. Now, to your question, you've done everything right up until when you calculate the uncertainty in the density. Suppose we have two quantities with their appropriate errors;

[tex](A\pm\delta A)\hspace{1cm}(B\pm\delta B)[/tex]

Now, suppose we want to find X = A/B, then the uncertainty in this case is;

[tex]\frac{\delta X}{X} = \sqrt{\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2}[/tex]

Also note that when you divide a value by a constant, you must also divide the error by that same constant. Hope this helps.
 
  • #4
I was a little unclear why the radius uncertainty is multiplied my three. Is there a relation between the exponent and the amt its multiplied by?? Truly wondering.
 
  • #5
denverdoc said:
I was a little unclear why the radius uncertainty is multiplied my three. Is there a relation between the exponent and the amt its multiplied by?? Truly wondering.
Indeed there is (as far as I know). If we again have some quantity [itex](A\pm\delta A)[/itex] and we wish to find [itex]Z = A^n[/itex] then the error in this case is;

[tex]\frac{\delta Z}{Z} = n\cdot\frac{\delta A}{A}[/tex]
 
  • #6
Thanks hoot. Learn something every day on this forum.
 
  • #7
denverdoc said:
Thanks hoot.
No problem :smile:
denverdoc said:
Learn something every day on this forum.
Oh how true that is, more often that not its me doing the learning!
 

1. What is the formula for determining the density of a sphere?

The formula for determining the density of a sphere is density = mass/volume, where mass is the amount of matter in the sphere and volume is the amount of space the sphere occupies.

2. How do you measure the mass of a sphere?

The mass of a sphere can be measured using a balance or scale. The sphere is placed on the balance and the measurement is recorded in grams (g) or kilograms (kg).

3. How do you calculate the volume of a sphere?

The volume of a sphere can be calculated using the formula V = (4/3)πr^3, where r is the radius of the sphere. Alternatively, the volume can be measured by submerging the sphere in a known volume of water and measuring the displacement.

4. Does the density of a sphere change with size?

No, the density of a sphere does not change with size. As long as the material and composition of the sphere remains the same, the density will remain constant regardless of the size of the sphere.

5. Can the density of a sphere be negative?

No, the density of a sphere cannot be negative. Density is a measure of how much mass is contained within a given volume and cannot have a negative value.

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