algebra III question about direct,inverse and joint variations?

Question

1.) the kinetic energy K varies jointly as the mass M and the square of the velocity V. if K is 30ergs when m is 8 gm and v is 3 cm/sec,find k if m=4 gm and v=6 cm/sec.





2.)the quantity Z varies directly as the square of X and inversely as the cube of y;z has the value 2/9 when x=2 and y=3. find the value of Z when X =1 and Y=4.






3.) The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. If a wire 50 m long and . 38cm in a diameter has a resistance of 3.5 ohms. what is the resistance of a wire of the same material 75m long and 0.625 cm in a diameter...




who ever answer those question will be rated as my best answer thank you!

Answer 1

I think this is what you are looking for:

(1)
Let E be for energy (to avoid confussion with k's).
Let α be a proprotional sign.
E α M*v² (from information given).
E = k M*v²
30 = k [(3x10^-3)*(3²)]
30 = k [72x10^-3]
30/ 72x10^-3 = k
k = 416∙666....
Now use the constant value of k to calculate your answer.
E = (416∙666....)(4x10^-3)(6²)
E = (416∙666....)(4x10^-3)(36)
E = 60 Joules of energy.

(2)
Z α x²(1/y^3)
Z = k [x²(1/y^3)]
2/9 = k [2²(1/3^3)]
2/9 = k [4(1/27)]
k = 2/9*27/4
k = 54/18
k = 3
Now use the constant value of k to calculate your answer.
Z = k [x²(1/y^3)]
Z = 3 [1²(1/4^3)]
Z = 3(1/64)
Z = 3/64

(3)
R α L*1/D²
R = k(L*1/D²)
3∙5 = k(50*1/(0∙38x10^-2)²
3∙5 = k(50*1/0∙00001444...)
3∙5 = k(50*69252∙07756)
3∙5 = k(3462603∙878)
3∙5 / (3462603∙878) = k
k = 0∙000 001 01
Now use the constant value of k to calculate your answer.
R = ( 0∙000 001 01..)(75*1/(0∙625x10^-2)²)
R = ( 0∙000 001 01..)(75*1/0∙000 039 062)
R = ( 0∙000 001 01..)(1920 000)
R = 1∙940736 Ω
R ≈ 1∙94 Ω

Answer 2

K is proportional to m/v^2,
so K= cm/v^2

Therefore c = Kv^2/m
= (substituting your original values) 30x9/8 = 270/8 = 135/4
(It's usually a good idea to find common factors, as smaller numbers are easier to handle than large.)

Substituting back into the original equation K = cm/v^2 using the new values gives
K = 135/4 x 4/36
= 45/9 = 5 (presumably the unit is ergs)

2. When I'm rearranging equations, I always find it easiest to "do the same thing to both sides". It may take a little longer than some of the rules you may have been taught, but it's more reliable.
So, putting the words into squiggles,
z is proportional to x^2/y^3 or
z = cx^2/y^3
Multiply both sides by y^3, gives
y^3 z = c x^2
Divide both sides by x^2, this gives
y^3 x/x^2 = c, or
c = y^3 z/x^2

So c = [(3^3) * (2/9)]/4
= 9*(2/9)/4
= 2/4 = 1/2.
So, when x = 1 & y = 4,

z = (1/2) * (1^2) /(4^3)
= [(1/2) *1]/64
= 1/128

3. Use a similar method to the above.
R is proportional to l/d^2, so
R = cl/d^2

Therefore c = Rd^2/l
Unlike the two problems above, the numbers are awkward in this instance, and so it is best to whack this number into a calculator, which gives

c = 0.010108.

Hence, when l = 75 & d = 0.625,

R = 1.293824 (unless I've made an arithmetical error somewhere!)

Answer 3

1) k=1/2mv^2= 1/2*4*6^2 = 1/2 * (8/2)* (3*2)^2 = 1/2*8*3^2*((1/2)*2^2) = 30*2 = 60 erg.

2) Z=kx^2/y^3 = k 1^2/4^3 = k (2/2 )^2 / (4*3/3)^3 = (k 2^2/ 3^3) * (1/2)^2/(4/3)^3 = (2/9)* (1/4) / (64/27) = (2/9) * (27/64*4) = 3/128

3) R=k l/d^2 = k 75/0.625^2 = k 75*(50/50)/(0.625*0.38/0.38)^2
= k*50/0.38^2 * (75/50)/(0.625/0.38)^2 = 3.5 * (3/2)/(6.5)^2
= i think you'll need a calc for this.