Understand T=k3 x H/N | Help & Explanation

[PainterGuy]

hello everyone,

suppose i have ("x" stands for multiplication)
1:-- T=k1 x H; and
2:-- T=k2 x 1/N;
3:-- then T=k3 x H/N

how do i combine "1" and "2" to get "3". trying to understand conceptually. help me please. many thanks for any help you can give me.

cheers

 

[HallsofIvy]

Your notation is not clear. I would interpret "k1" and "k2" as different values of k indexed by 1 and 2 but apparently you mean powers: k^1= k and k^2= k*k
\left(k H\right)\left(\frac{k^2}{N}\right)= \frac{k(k)(k) H}{N}= \frac{k^3H}{N}= k^3\frac{H}{N}

 

[PainterGuy]

hello everyone,

suppose i have ("x" stands for multiplication) and k₁, k₂, and k₃ are any constants:
1:-- T=k₁ x H; and
2:-- T=k₂ x 1/N;
3:-- then T=k₃ x H/N

how do i combine "1" and "2" to get "3". trying to understand conceptually. help me please. many thanks for any help you can give me.

cheers

Your notation is not clear. I would interpret "k1" and "k2" as different values of k indexed by 1 and 2 but apparently you mean powers: k^1= k and k^2= k*k
\left(k H\right)\left(\frac{k^2}{N}\right)= \frac{k(k)(k) H}{N}= \frac{k^3H}{N}= k^3\frac{H}{N}

many thanks HallsofIvy. my quoted post above is edited to reflect what i actually wanted to say. why are you multiplying "1" and "2" to get "3".

i understand in such proportional relations multiplication is done but why? why don't we add?

cheers

 

[KrisOhn]

many thanks HallsofIvy. my quoted post above is edited to reflect what i actually wanted to say. why are you multiplying "1" and "2" to get "3".

i understand in such proportional relations multiplication is done but why? why don't we add?

cheers

If I understand you correctly, we do add. (k^{1})(k^{2}) = (k^{3})

This can be explained by (k)(k \cdot k) = k \cdot k \cdot k = k^{1+1+1} = k^{3}

The only time we multiply exponent cases is when we have something like (k^{2})^{3} = k^{6}

 

[PainterGuy]

If I understand you correctly, we do add. (k^{1})(k^{2}) = (k^{3})

This can be explained by (k)(k \cdot k) = k \cdot k \cdot k = k^{1+1+1} = k^{3}

The only time we multiply exponent cases is when we have something like (k^{2})^{3} = k^{6}

many thanks KrisOhn.

my post# 3 is corrected. there is no k³, k², etc. now help me please

cheers

 

[KrisOhn]

You're going to have to rephrase your question in a clearer manner, I don't see what you're getting at other than what we've already outlined.

 

[PainterGuy]

You're going to have to rephrase your question in a clearer manner, I don't see what you're getting at other than what we've already outlined.

okay here it is.

Resistance is proportional to length: R \propto L
Resistance is inversely proportional to Area: R \propto 1/A

1:- R = k₁ x L
2:- R = k₂ x 1/A
3:- R = \rho L/A

k₁ and k₂ are constants of proportionality.

how do we get "3" from "1" and "2". tell me please now. many thanks.

cheers

 

[PainterGuy]

hello,

will someone please help me out? i will be grateful. if something is still unclear please tell me. i will try to clear it up.

cheers

 

[uart]

okay here it is.

Resistance is proportional to length: R \propto L
Resistance is inversely proportional to Area: R \propto 1/A

1:- R = k₁ x L
2:- R = k₂ x 1/A
3:- R = \rho L/A

k₁ and k₂ are constants of proportionality.

how do we get "3" from "1" and "2". tell me please now. many thanks.

cheers

Because k_1 = \rho / A and k_2 = \rho L

That is, k1 is only a constant if A is fixed and k2 is only constant if L is fixed. What I mean is that R = k_1 L is only valid for the case where "L" is the only thing being varied.