~it was bumped over to the next page so I recopied it and added my answer for the question and was wondering if my integration was correct since you said
DrWls:I forgot t use the expression B + Cx for p, and used A + Bx instead. The method of solution is the same. In your case, B = 2.70, and C = 1.186
p(x) = 2.70 + 1.186 x
Note that p = 2.70 g/cm^3 when x = 0 and 9.30 g/cm^3 when p = 14 cm, as required.
Now integrate A p(x) dx for the total mass
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Q--A rod extending between x=0 and x= 14.0cm has a uniform cross-
sectional area A= 9.00cm^2. It is made from a continuously changing
alloy of metals so that along it's length it's density changes
steadily from 2.70g/cm^3 to 19.3g/cm^3.
a) Identify the constants B and C required in the expression p= B +
Cx to describe the variable density.
b) The mass of the rod is given by
m= integral(pdV)= integral(pAdx)= integral 14cm/0 (14cm above integral sign and 0 below integral)(B+Cx)*(9.00cm^2)dx
(below the firs integral it says all material and below the
second integral it says all x
~for this I got
m= A integral 14\0 (B+Cx)dx
(took out the A contstant A=9.00cm^2)
m= 9.00 integral 14\0 (2.70+1.186x)dx =
9.00(2.70x + 0.593x^2)|14 0 =
(the 14 is above and 0 below the | sign)
9.00[ 2.70(14) + 0.593(14)^2 ]=
1,386.252g
~I'm not sure if I did the integration right and also I'm confused as to why you said
"note that p= 2.70g/cm^3 when x=0 and 9.30g/cm^3 when p= 14cm, as required" does this apply to what I plug into the integration ?? I would think it wouldn't since I have to put in 14 and 0 in anyways and the 0 cancels out and all is left is the 14 right?
Is this alright what I got??
6 answers
THANK YOU =D
1 answer