What is the density of pure chromium in g cm^-3?

[Richie Smash]

Yes I agree... the exam I'm writing isn't actually a high level, and the book even has harder questions than the exam itself, I agree I'm not understanding some simple algebraic things here... I apologize.

What I understand is up to the second line in my attempt in post 25, and how you got r = D2/4D1 = V1/V2 in terms of the equation D2V2 = 4 D1 V1.
I don't understand how substituting r into the second line of my post 25, i,e D3 = (D1V1+4D1V1)/V1+V2 turns into 5 D1 r/ r+1.

I'm writing the general maths exam as well, the algebra in it is not too complicated, so I don't have experience in these things, but thank you for helping me out Charles and taking the time to guide me, I am going to put in hours and hours in the coming months, but right now I need to spend a long time understanding what you're trying to do, my problem is understanding the simple steps, like ''why does doing this get rid of these terms?'' or '''why is this the ratio of this'' etc.

 

[Charles Link]

The first thing you need is to take the equation that is of the form $D3=\frac{A}{B}$ and divide numerator and denominator by $V2$: $D3=\frac{(\frac{A}{V2})}{(\frac{B}{V2})}$. Then process each part $\frac{A}{V2}$ and $\frac{B}{V2}$ separately. $\\$ You should also have recognized that $D1 \, V1 +4 \, D1 \, V1 =5 \, D1 \, V1$. Doing the complete algebra: $(1) (D1 \, V1)+ (4) (D1 \, V1)=(4+1)(D1 \, V1)=5 \, D1 \, V1$ . $\\$ For the above $A=5 \, D1 V1$, and $B=V1+V2$. It should be a simple matter to divide $A$ by $V2$, and $B$ by $V2$, and then substitute in for $r=\frac{V1}{V2} =\frac{D2 }{4 \, D1}$.

 

[Richie Smash]

Ok I understand how you got A = 5D1V1 and B = V1+V2 {V2 = (4D1V1)/D2
B divided by V2 = V1
and A divided by V2 = 5D1V1 x (D2)/(4D1V1)
= 5D2/4
So the you would have (5D2/4)/V1 {V1 = (D2V2)/4D1}

Now that works out to (5D2)/4 x (4D1)/(D2V2)
= (20D1)/4V2
= 5D1/V2.

Am I getting somewhere?

 

[Charles Link]

Ok I understand how you got A = 5D1V1 and B = V1+V2 {V2 = (4D1V1)/D2
B divided by V2 = V1
and A divided by V2 = 5D1V1 x (D2)/(4D1V1)
= 5D2/4
So the you would have (5D2/4)/V1 {V1 = (D2V2)/4D1}

Now that works out to (5D2)/4 x (4D1)/(D2V2)
= (20D1)/4V2
= 5D1/V2.

Am I getting somewhere?

Progress is somewhat slow. $B$ divided by $V2$ is $\frac{B}{V2}=1+\frac{V1}{V2}$ and then you need to substitute in for the expression for $\frac{V1}{V2}$.

 

[Richie Smash]

Sorry for taking up so much of your time.
I thought that since B=V1+V2 then dividing by V2 would get rid of the two terms and leave V1.
I'm not quite sure how you arrived at your answer, but I assume I solved A wrong as well?

 

[Charles Link]

Sorry for taking up so much of your time.
I thought that since B=V1+V2 then dividing by V2 would get rid of the two terms and leave V1.
I'm not quite sure how you arrived at your answer, but I assume I solved A wrong as well?

Yes, the numerator with $A$ is also incorrect. One of the Physics Forums rules in the Homework section is that the Homework Helper is not allowed to furnish a complete solution. The student is expected to do most of the work, with the Homework Helper, for the most part, simply giving helpful hints. Your algebra background is rather deficient, and you really need to get semi-proficient at it before you can expect to have any kind of success with the physics problems that you are attempting to solve.

 

[Richie Smash]

OK, I understand how you've gotten B = 1+ D2/(4D1)
Now I'm going to learn how you've gotten A = 5D1 *r.

Ok I understand exactly how you got A = 5D1 (V1/V2)

Now I'm going to substitute in D2/4D1 and see what happens.

 

[Charles Link]

OK, I understand how you've gotten B = 1+ D2/(4D1)
Now I'm going to learn how you've gotten A = 5D1 *r.

Just an observation=there seems to be a considerable difference between any mathematical skills that you may have, compared with your verbal skills. Some people simply struggle with mathematics. If that is the case, I encourage you to keep trying, but if you did not pick up some of the mathematics from early on, it may be very slow going. From some of your previous posts, your verbal skills are considerably above average, perhaps even considerably above average among college students. But if the algebra is as slow as it appears, I do recommend you invest in a private tutor if you want to learn it. A chalk board and/or pencil and paper is almost a necessity. The Latex on the computer is not properly equipped to easily show any algebraic steps.

 

[Richie Smash]

Thank you for the continued advice Charles, I shall come clean, currently I'm in a special situation. I had to drop out of school a while back, and for many years I was not schooled, I believe I missed out on a lot of concepts, but once I understand something I do, at the moment I am considering getting a tutor, but for the while I've been progressing by just reading trial and error, and I've definitely improved from some months ago, I'm putting in a tremendous amount of work atm and my slow progress pays off, because a few months ago there were certain things I couldn't do quickly, now I can.

Btw, was your answer 1.712?

 

[PAllen]

Thank you for the continued advice Charles, I shall come clean, currently I'm in a special situation. I had to drop out of school a while back, and for many years I was not schooled, I believe I missed out on a lot of concepts, but once I understand something I do, at the moment I am considering getting a tutor, but for the while I've been progressing by just reading trial and error, and I've definitely improved from some months ago, I'm putting in a tremendous amount of work atm and my slow progress pays off, because a few months ago there were certain things I couldn't do quickly, now I can.

Btw, was your answer 1.712?

If that is meant to be the density of Chromium under the 80/20 mass assumption, no, it is not close to correct.

I do continue to think the approach I described in post #21 would likely have been simpler for you.

 

[Richie Smash]

If that is meant to be the density of Chromium under the 80/20 mass assumption, no, it is not close to correct.

I do continue to think the approach I described in post #21 would likely have been simpler for you.

I haven't forgotten about that one either ;), don't worry I actually want to solve both ways at a time, to be honest I was trying your way and i got stuck as well haha.
But I'm progressing none the less.

 

[Richie Smash]

OK, So I'm making progress.

D3 = (5D1V1)/(V1+V2)
Divide both numerator and denominator by V1/V2

I'll get D3 = (5D1(V1/V2))/(1+V1/V2)

Making D1 the subject I'll get D1 = (D3(1+V1/V2)) / (5(V1/V2))

This can be simplified by dividing by V1/V2 again

I'll get D1 = D3/5 * (V2/V1+1)

And now I can use my value r which = D2/(4D1)

D1= D3/5 *((4D1)/D2 +1)

 

[Charles Link]

OK, So I'm making progress.

D3 = (5D1V1)/(V1+V2)
Divide both numerator and denominator by V1/V2

I'll get D3 = (5D1(V1/V2))/(1+V1/V2)

Making D1 the subject I'll get D1 = (D3(1+V1/V2)) / (5(V1/V2))

This can be simplified by dividing by V1/V2 again

I'll get D1 = D3/5 * (V2/V1+1)

And now I can use my value r which = D2/(4D1)

D1= D3/5 *((4D1)/D2 +1)

If you divide both sides of this equation by $D1$, you have, with one more algebraic step, the result from @PAllen of post 23. You can then solve for $D1$, and plug in the numbers. I strongly recommend getting a tutor for learning algebra, because it is too difficult to teach over the internet. $\\$ There is one tip I can give you that much of algebra is based on: The equation is like a balance where you weigh things. You can add or subtract the same weight on both sides of the equal sign and it stays balanced. You can multiply or divide what is on both sides of the equal sign by the same amount and it stays balanced. The rest is simply getting the terms that you want on one side of the equal sign, and the rest on the other. $\\$ Learning physics seems to work well over the internet, but algebra really needs pencil and paper, and/or a chalk board.

 

[Richie Smash]

Right, so I have

D1= D3/5 *((4D1)/D2 +1)
Divide both sides by D3/5.

I'll get (5D1)/D3 = ((4D1)/D2 +1)

Divide both sides by D1
I'll get 5/D3 = 4/D2+1/D1

Subtract 4/D2 from both sides

And now I finally have the final expression.

I have 1/D1 = 5/D3 - 4/D2

From here I solved for D1 and obtained 7.4gcm^-3 as the density of D1.

 

[PAllen]

And that is correct.