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. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  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. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  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. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. 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. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  4. and since p=m/V does that play a role in this as well?

    Again I have no idea what C represents.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  5. 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. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  6. 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. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Similar Questions

  1. physics

    A net charge of 47 mC passes through the cross-sectional area of a wire in 19.0 s. (a) What is the current in the wire? 1 A (b) How many electrons pass the cross-sectional area in 1.0 min? 2 . electrons I got 9.3e+22 for B but its

  2. 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

  3. Fluid Machineries

    Water is flowing in a pipe with a circular cross-section area, and at all points the water completely fills the pipe. At point 1, the cross-section area of the pipe is 0.080-m2 and the fluid velocity is 4.50 m/s. (a) What is the

  4. 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?

  1. Hydraulic Physics

    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? Would I need to divide both pistons areas

  2. Physics

    A hydraulic lift in a garage has two pistons: a small one of cross-sectional area 3.70cm2 and a large one of cross-sectional area 210cm2 . Part A If this lift is designed to raise a 3100-kg car, what minimum force must be applied

  3. 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

  4. physics

    water flows along a horizontal pipe of cross-sectional area 48cm^2 which has a constriction of cross-sectional area 12cm^2 at one place. If the speed of the water at the constriction is 4 m/s, calculate the speed in the wider

  1. physics

    A tube of uniform cross sectional area closed at one end contain some dry air which is sealed by a thread of Mercury 15.5cm long.when the tube is held vertically with the closed end at the bottom the air column is 22.0cm long, but

  2. 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

  3. physics

    An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q = 1.0μC, is located at the point (x, y, z) = (0.00cm, 1.0cm, 0.00cm), while the negative charge is located at the point (x,

  4. physics

    A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.849 m and 1.17 kg. What constant-magnitude force acting at the other end of the rod perpendicularly

View more similar questions or ask a new question.