Calculating Mass of Sand Grains with Given Surface Area for Physics Problem

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In summary, we need to use the equation for surface area of a sphere, A = 4πr^2, and the equation for volume, V = (4/3)πr^3. We also need to use the equation for mass, M = Vρ, where ρ is the density. After plugging in the given values, we find that the mass of sand grains with a total surface area equal to a 1.0 m cube is approximately 0.25 kg. It is normal to struggle with problems in a physics class, but with practice and determination, you can improve your understanding and succeed in the course.
  • #1
Gattz
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Homework Statement


Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.5 × 10^3 kg/m^3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.0 m on an edge?


Homework Equations


I think I need to use Surface area of sphere A=4pir^2, but no equation was given.


The Attempt at a Solution


First I did 4x3x(5e-8)^2 = 3e-16m^2 to find surface area of individual sphere. Then I got 6 by doing surface area of a cube = 6(1)^2. I divided 6m^2 by 3e-16m^2 (Correct me I'm even getting the units right which I don't believe so). Then I got 2e16 (units?). Finally I multiply that by 2.5e3 kg/m^3 and get 5e19. All I know is that the units for final answer is kg, so m^3 gets canceled out some how.

P.S. This is for my general physics calc based class and I feel extremely dumb... If I am having trouble with these problems is that an indicator that I shouldn't be taking this class? I've taken Calc 1 at my community college a year ago, and I did well in that, but now I'm taking Calc 2 and this and I feel like I don't know anything. I don't have a major, but I hear physics course is good to take no matter. This and the algebra physics are basic physics courses so if I'm not doing well here does that mean I can't do physics?
 
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  • #2
Gattz said:

Homework Statement


Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.5 × 10^3 kg/m^3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.0 m on an edge?


Homework Equations


I think I need to use Surface area of sphere A=4pir^2, but no equation was given.


The Attempt at a Solution


First I did 4x3x(5e-8)^2 = 3e-16m^2 to find surface area of individual sphere. Then I got 6 by doing surface area of a cube = 6(1)^2. I divided 6m^2 by 3e-16m^2 (Correct me I'm even getting the units right which I don't believe so). Then I got 2e16 (units?). Finally I multiply that by 2.5e3 kg/m^3 and get 5e19. All I know is that the units for final answer is kg, so m^3 gets canceled out some how.

P.S. This is for my general physics calc based class and I feel extremely dumb... If I am having trouble with these problems is that an indicator that I shouldn't be taking this class? I've taken Calc 1 at my community college a year ago, and I did well in that, but now I'm taking Calc 2 and this and I feel like I don't know anything. I don't have a major, but I hear physics course is good to take no matter. This and the algebra physics are basic physics courses so if I'm not doing well here does that mean I can't do physics?

You should be fine in class. It takes a bit of problem solving to get the hang of it, so just keep on working problems.

Hint #1 -- you need to use: mass = volume * density That is where the m^3 term is handled.

Hint #2 -- there was a very similar thread about this problem a few days ago in this forum...
 
  • #3
Okay, I tried doing M=VxD. V=(4/3)pi(5e7)^3 I got the 5e7 by doing 50um/10^-6. So volume is now 5.24e23 m^3. Now I multiplied that by the density (I see that the m^3 cancels), and I get 1.31e27 kg. Now I'm very confused... A grain of sand weighing that much?? What did I do wrong?

Edit - I see what I did. I did 50/10^-6 instead of 50/10^6.
 
  • #4
Your value for the radius of the sand grains appears to be off by a factor of 1000.
 
  • #5
Wait what do you mean? I got 5e-5 meters as the radius. You mean it's suppose to be 5e-8 or 5e-2?
 
  • #6
Gattz said:
Wait what do you mean? I got 5e-5 meters as the radius. You mean it's suppose to be 5e-8 or 5e-2?

50um = 50*10^-6m = 5*10^-?m
 
  • #7
BTW, you also used 3 for PI in the original post. You should at least use 3.14.
 
  • #8
5e-5 m.
 
  • #9
Correct.
 
  • #10
Now found surface area 3.14e-8m^2 and divided 1m by that. So that would turn out to be 3.18e7/m. Then I took that multiplied by 6 and got 1.91e8 kg. And got .25kg. Is this right or wrong?

Okay that seems to be right, but can someone go over the units for what I just did? I think I got lost somewhere along that.
 
  • #11
Gattz said:
Wait what do you mean? I got 5e-5 meters as the radius. You mean it's suppose to be 5e-8 or 5e-2?

You wrote 5e-8; maybe just a typo.
 
  • #12
Oh yeah, in the first post my work was all wrong.
 

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