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Math Help - Having real trouble with this question...

  1. #1
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    Having real trouble with this question...

    Hello everyone... I am in a university-level science coarse and I am having real trouble with this question on our HW assignment as I am not a math oriented individual.... Any help is greatly appreciated. I know its long, but I am really terrible at this whole math thing.

    Thanks again..



    Their starting point was the assumption that the total mass or iridium in the K/T
    layer was equal to the mass of iridium in the original impactor:
    m
    Ir(layer) = mIr(impactor)

    1) First you must estimate the mass of iridium in the K/T layer. Use the following
    information:
    a. On average, the iridium layer is
    H = 3 cm thick (what is it in SI units?).
    b. The density of the K/T layer containing iridium is
    d = 2.5 g/cm3 (what is it in SI
    units?).
    c. On average, the layer had a concentration of iridium of
    CIr(layer) =20 parts per
    billion (20 ppb) by weight. (
    Hint: in scientific notation, 1 billion is 109, so 1 ppb is
    equivalent to 1/10
    9 or 10-9)
    d. Assume that the iridium concentration was uniform around the Earth.
    e. The total area of the Earth is
    AEarth = 4 π(REarth)2

    where the radius of Earth is
    REarth = 6378 km (what is it in SI units?).
    f. The total mass of the K/T layer,
    m(layer), is given by the area of the Earth (AEarth)
    multiplied by the layer thickness (
    H) multiplied by the layer density (d).
    g. The total mass of iridium in the K/T layer is given by the mass of the layer
    multiplied by the iridium concentration:

    m
    Ir(layer) = m(layer)CIr(layer)


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  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
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    Quote Originally Posted by fawd View Post
    Hello everyone... I am in a university-level science coarse and I am having real trouble with this question on our HW assignment as I am not a math oriented individual.... Any help is greatly appreciated. I know its long, but I am really terrible at this whole math thing.

    Thanks again..



    Their starting point was the assumption that the total mass or iridium in the K/T
    layer was equal to the mass of iridium in the original impactor:

    m
    Ir(layer) = mIr(impactor)

    1) First you must estimate the mass of iridium in the K/T layer. Use the following
    information:
    a. On average, the iridium layer is H = 3 cm thick (what is it in SI units?).
    b. The density of the K/T layer containing iridium is d = 2.5 g/cm3 (what is it in SI
    units?).
    c. On average, the layer had a concentration of iridium of CIr(layer) =20 parts per
    billion (20 ppb) by weight. (Hint: in scientific notation, 1 billion is 109, so 1 ppb is
    equivalent to 1/109 or 10-9)
    d. Assume that the iridium concentration was uniform around the Earth.
    e. The total area of the Earth is AEarth = 4 π(REarth)2
    where the radius of Earth is


    REarth = 6378 km (what is it in SI units?).

    f. The total mass of the K/T layer, m(layer), is given by the area of the Earth (AEarth)
    multiplied by the layer thickness (H) multiplied by the layer density (d).
    g. The total mass of iridium in the K/T layer is given by the mass of the layer
    multiplied by the iridium concentration:
    m


    Ir(layer) = m(layer)CIr(layer)
    First calculate the mass if the iridium layer. This is A\times d \times \rho where A is the surface area of the Earth and d is the thickness of the iridium layer and \rho the density of the layer. All of these should be in SI units, that is A is in square metres, d in metres, and \rho in kg/cubic metre.

    <br />
A=4 \pi R^2<br />

    where R is the radius of the Earth (in metres).

    CB
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