1. Questions

Physics, Calculus(alot of stuff together)= HELP!!

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

~this part a has gotten me confused as to what to do at all and the second part has gotten me even more confused!!

HELP!!!

Thank YOU!

  1. 👍 0
  2. 👎 0
  3. 👁 1,295
~christina~

  1. I forgot to include what it says under the integration:

    Carry out the integration to find the mass of the rod

    Now about that..what in the equation for the integral is the mass???
    it's not like they have a m anywhere in there..unless it's p but I started thinking it was to represent density..

    1. 👍 0
    2. 👎 0
    ~christina~

  2. Yes, p represents density in this case. The mass of the rod is
    A*(Integral of) p (x) dx
    where A is the cross sectional area of the rod.
    For p(x), use the function you derive in part (a).
    p(x) = A + bx
    p(0) = A = 2.70 g/cm^3 (That already tells you what A is)
    p(14) = A + B*14 = 19.3 g/cm^3
    Solve for B.

    1. 👍 1
    2. 👎 0
    drwls

  4. first of all I don't know how to use the #s given to put into the equation or do something else with them.

    I know...

    x=0
    x=14.0cm
    A=9.00cm^2

    density changes from: 2.70g/cm^3 to 19.3g/cm^3

    How do I put it into p= B+Cx??

    would x= 14-0??
    C and B I'm not sure what they represent except that you said the mass is equal to B right?

    1. 👍 1
    2. 👎 0
    ~christina~

  5. and since p=m/V does that play a role in this as well?

    Again I have no idea what C represents.

    1. 👍 0
    2. 👎 0
    ~christina~

  6. 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

    1. 👍 0
    2. 👎 0
    drwls

  8. 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??

    1. 👍 0
    2. 👎 0
    ~christina~

Respond to this Question

First Name

Your Response

Similar Questions

  1. PHYSICS !!!

    The diffusion rate for a solute is 4.0 10-11 kg/s in a solvent-filled channel that has a cross-sectional area of 0.50 cm2 and a length of 0.220 cm. What would be the diffusion rate m/t in a channel with a cross-sectional area of

  2. physics

    The intake in the figure has cross-sectional area of 0.75 m2 and water flow at 0.41 m/s. At the outlet, distance D = 180 m below the intake, the cross-sectional area is smaller than at the intake and the water flows out at 9.6

  3. Physical Education

    Suppose that blood flows through the aorta with a speed of 0.35 m/s. The cross-sectional area of the aorta is 2.0x10-4 m2. (a) Find the volume flow rate of the blood (b) The aorta branches into tens of thousands of capillaries

  4. Physics

    A liquid (density = 1.65 g/cm^3) flows through two horizontal sections of tubing. In the first section the cross-sectional area is 10.0 cm^2, the flow speed is 275 cm/s, and the pressure is 1.20 x 10^5 Pa. This tubing then rises

  1. physics

    A venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed v2 through a horizontal section of pipe whose cross-sectional area A2 = 0.0900 m2. The gas has a density of

  2. Physics please help!!

    Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.50·10-2 m2, the pressure is 5.30·105 Pa and the velocity is 0.310 m/s. In a constricted region where the

  3. physics

    Two conductors are made of the same material and have the same length. They each have circular cross sectional areas. The ratio of their radii is 2:1. they carry the same current. What is the ratio of the electrical potential

  4. Fluid Mechanics

    In a hydraulic system a 20.0-N force is applied to the small piston with cross sectional area 25.0 cm2. What weight can be lifted by the large piston with cross sectional area 50.0 cm2?

  1. Physics

    Water enters a pipe with a cross sectional area of 3.0 cm^2 with a velocity of 3.0 m/s. The water encounters a constriction where its velocity is 15 m/s. What is the cross sectional area of the constricted portion of the pipe?

  2. Precal

    The velocity v of a fluid flowing in a conduit is inversely proportional to the cross- sectional area of the conduit.( Assume that the volume of the flow per unit of time is held constant .) Determine the change in the velocity of

  3. Electricity

    50.0pJ of energy is stored in a 3.00cm × 3.00cm × 3.00cm region of uniform electric field. What is the strength of the field?

  4. Physics

    Two wires have the same length and the same resistance but are made up of different materials. The resistivities of these materials are 3.97 Ù and 1.97 Ù. Obtain the ratio of the smaller cross-sectional area to the greater

You can view more similar questions or ask a new question.