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Units of a Gradient on a graph?

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    Imm trying to work out the gradient of agraph, but i need the unit of the gradient, but the y axis is in logs. Does this change the unit in any way?.
    E.g the y axis is ln(V), V being voltage measured in V, x axis is volume of liquid, measured in ml. What is the unit of the gradient?
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    (Original post by megalolz92)
    Imm trying to work out the gradient of agraph, but i need the unit of the gradient, but the y axis is in logs. Does this change the unit in any way?.
    E.g the y axis is ln(V), V being voltage measured in V, x axis is volume of liquid, measured in ml. What is the unit of the gradient?
    I have a question first. Did you take the logarithm of voltage to get a straight line on the graph? Does it go through the origin?

    I am asking because if it is a straight line, and \ln u=\alpha V where u is voltage, \alpha is the gradient and V is volume, you obtain the following: u=\mathrm{e}^{\alpha V}. The left hand side side has a dimension of volts, the right hand side is dimensionless... something doesn't work here. Are you sure it's not like this: u=u_0\, \mathrm{e}^{\alpha V} where u_0 is some initial value of voltage? (Then the straight line would be \ln \tfrac{u}{u_0}=\alpha V.)

    If this is an experiment, can you briefly describe it?
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    Yeah it's like an physics assessment experiment that i did. There is a solar cell below a beaker filled with ink, connected to a voltmeter. When plotted on a graph it gives a straight line with negative gradient. I found the gradient and put the unit of it as ml^-1 because ln(V) doesn't have a unit. Not sure if it's right but i can't change it now anyway.
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    That should be ok for the gradient in this case if log V is on the vertical/y axis and x on the horizontal.
    The gradient gives the value of the constant c in
    V= ke-cx
    and as jaroc says, c will have units of inverse volume because -cx must be dimensionless.
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    Ln(V) is just a number, so your gradient will be in terms of ml^-1 or per ml.
    I had a similar issue doing my electro chemistry coursework for A2 chemistry. It makes sense that it should just be ml^-1 (or V^-1 in my case), but it's always good to check that that's the done thing.
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    I have a question about units of a gradient. If I am plotting Newtons Metres (Nm) against Metres (m), then the units for my gradient would be Nm/m, i.e. N, Newtons. However, in the practical skills handbook on the OCR website, its specifically says:
    "e. A gradient value has no unit since it is a ratio of two numbers from the graph"

    Who is correct?
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    (Original post by geditor)
    I have a question about units of a gradient. If I am plotting Newtons Metres (Nm) against Metres (m), then the units for my gradient would be Nm/m, i.e. N, Newtons. However, in the practical skills handbook on the OCR website, its specifically says:
    "e. A gradient value has no unit since it is a ratio of two numbers from the graph"

    Who is correct?
    As far as the OCR booklet is concerned it depends what graph they are talking about. Some graphs have a gradient with no units/dimensions.
    In the case of the one you mention, if it's Nm on the y axis and m on the x axis, the gradient is y/x which is Nm/m which gives N

    If the OCR booklet is saying that all graphs have a dimensionless gradient I would want to see the complete context of this statement before passing any judgement.
    It depends what they define as "two numbers" from the graph.
    It sounds like they may be referring purely mathematically to the gradient.
    In practical physics your graph can quite acceptably have a gradient with a unit.
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    The principle is the we only plot numbers on a graph

    eg if L = 5m then we plot L/m which is 5
    and if m=0.1kg we plot m/kg which is also a pure number = 0.1

    Thats the idea of labelling table headings and axes as quantity divided by unit.

    With this convention everything plotted is a number so the gradient has no units since its a number divided by a number.
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    Yes, strictly speaking, you only plot numbers. That's the problem here.
    The point is that the (implied) units of the quantities plotted on the axes give you the (implied) units of the quantity represented by the gradient.
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    I know - either way its a horrible cludge.
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    Thanks

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Updated: February 1, 2012
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