[mechanics] radius of a sphere with critical mass

[Rijad Hadzic]

Homework Statement

Here is the question:

"In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a chain reaction. This element has a density of 19.5 g/cm3 .

What would be the radius of a sphere of this material that has a critical mass?

r = ______ cm ? "

So my answer should be in CM, and I should express my answer using 2 significant figures.

Homework Equations

Here, equations I used were

V = m * d

The Attempt at a Solution

What I've done so far:

Ok you have

Mass = 60 kg

D = 19.5g/cm3

I converted 60 kg to 60,000 g because D is in terms of g/cm3 and my final answer is suppose to be cm, so I think this helps simplify it.

anyways, I have the equation:

V = m * d

so

(4/3)pi*r3 = 60,000 g * 19.5g/cm3

Okay, I'm unsure what happens when I multiply 60,000 g by 19.5g/cm3

Does my result become 1170000 g2/cm3? or does it remain 1170000 g/cm3 ?

So from what I understand after this, I'm going to get

r3 = (3 (result of the above multiplication) ) / 4pi

and then take that result and 1/3 it to get my final answer.

I'm mostly stuck here on the multiplying 60,000 g by 19.5g/cm3 part.

I have the feeling that I understand how to do it but I actually can't do it :/!

 

[SteamKing]

Homework Statement

Here is the question:

"In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a chain reaction. This element has a density of 19.5 g/cm3 .

What would be the radius of a sphere of this material that has a critical mass?

r = ______ cm ? "

So my answer should be in CM, and I should express my answer using 2 significant figures.

Homework Equations

Here, equations I used were

V = m * d

The Attempt at a Solution

What I've done so far:

Ok you have

Mass = 60 kg

D = 19.5g/cm3

I converted 60 kg to 60,000 g because D is in terms of g/cm3 and my final answer is suppose to be cm, so I think this helps simplify it.

anyways, I have the equation:

V = m * d

so

(4/3)pi*r3 = 60,000 g * 19.5g/cm3

Okay, I'm unsure what happens when I multiply 60,000 g by 19.5g/cm3[\QUOTE]

Does my result become 1170000 g2/cm3? or does it remain 1170000 g/cm3 ?

Really? It should be very clear that multiplying grams × grams / cm³ ≠ cm³

Figure out how to get only cm³ [Hint: multiplication isn't the only thing you can do here.]

 

[Rijad Hadzic]

Really? It should be very clear that multiplying grams × grams / cm³ ≠ cm³

Figure out how to get only cm³ [Hint: multiplication isn't the only thing you can do here.]

Sorry, what I meant was, when multiplying grams by grams/cm^3, is my answer going to be grams^2/cm^3 or is it going to stay grams/cm^3?

Also, do you mean dividing?

If so, I tried the following:

60,000 g x cm^3 / 19.5 g to cancel out the g's, resulting with 3076.923077 cm^3.

After that I got ( (3076.923077 cm^3.) x 3 ) / 4pi = 7249.829201 cm^3so now I take (7249.829201 cm^3)^1/3 = 19.35423112 .

see, now I'm not sure what happened to the cm^3. when I did ^1/3, did that get rid of the ^3 and only make it cm? Or does it stay cm^3?

 

[SteamKing]

Sorry, what I meant was, when multiplying grams by grams/cm^3, is my answer going to be grams^2/cm^3 or is it going to stay grams/cm^3?

Also, do you mean dividing?

If so, I tried the following:

60,000 g x cm^3 / 19.5 g to cancel out the g's, resulting with 3076.923077 cm^3.

After that I got ( (3076.923077 cm^3.) x 3 ) / 4pi = 7249.829201 cm^3so now I take (7249.829201 cm^3)^1/3 = 19.35423112 .

see, now I'm not sure what happened to the cm^3. when I did ^1/3, did that get rid of the ^3 and only make it cm? Or does it stay cm^3?

Remember your algebra!

Density is defined as ##ρ = \frac{M}{V}##. Obviously, if you are looking for V, then you must manipulate this equation algebraically to obtain ##V = \frac{M}{ρ}##.
The units are manipulated in a similar fashion: ##\frac{g}{cm^3}=g\div cm^3##, so doing a similar manipulation to the one above, ##cm^3=\frac{g}{\frac{g}{cm^3}}##, which can be simplified by inverting the units in the denominator and multiplying by the numerator, ##\frac{g}{\frac{g}{cm^3}} = g × \frac{cm^3}{g}=cm^3##.

You get a volume of ##V = 3076.9\: cm^3 = \frac{4πr^3}{3}##.

When you solve for r³, the units are still cm³. By taking the cube root of r³ in cm³, you are left with the radius in cm.

 

[Rijad Hadzic]

Remember your algebra!

Density is defined as ##ρ = \frac{M}{V}##. Obviously, if you are looking for V, then you must manipulate this equation algebraically to obtain ##V = \frac{M}{ρ}##.
The units are manipulated in a similar fashion: ##\frac{g}{cm^3}=g\div cm^3##, so doing a similar manipulation to the one above, ##cm^3=\frac{g}{\frac{g}{cm^3}}##, which can be simplified by inverting the units in the denominator and multiplying by the numerator, ##\frac{g}{\frac{g}{cm^3}} = g × \frac{cm^3}{g}=cm^3##.

You get a volume of ##V = 3076.9\: cm^3 = \frac{4πr^3}{3}##.

When you solve for r³, the units are still cm³. By taking the cube root of r³ in cm³, you are left with the radius in cm.

Thank you so much :)! Just as you wrote this I finally got the answer! I appreciate it so much!