// Numbas version: exam_results_page_options {"name": "3: Quadratic Equations - Temperature vs. Depth", "extensions": [], "custom_part_types": [], "resources": [["question-resources/2016-08-19.png", "/srv/numbas/media/question-resources/2016-08-19.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

"}, "statement": "

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The temperature within the Earth's interior increases with depth to the centre of the earth (with radius, r).

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To find the temperature at a certain depth within the interior of the Earth we can use the following equation:

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$T=(-8.225\\times10^{-5})z^2+1.05z+1110$

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where;
$T$ = Temperature ($^\\circ{C}$)
$z$ = Depth ($km$)

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This equation gives a quadratic graph showing the relationship between temperature and depth.

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You should obtain two values for depth for each temperature because the depth is relative to which side of the earth you are considering the surface to be, as the earth is symmetrical. This is illustrated in the above figure, where X and Y are two values for depth for one given point at which temperature = $T^\\circ{C}$.

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Use the above equation to complete the statements below.

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$a)$ At $\\var{t1}^\\circ{C}$, the depth is [[0]]$km$ away from the closest surface and [[1]]$km$ away from the furthest surface

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$b)$ At $\\var{t2}^\\circ{C}$, the depth is [[2]]$km$ away from the closest surface and [[3]]$km$ away from the furthest surface

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$c)$ At $\\var{t3}^\\circ{C}$, the depth is  [[4]]$km$ away from the closest surface and [[5]]$km$ away from the furthest surface.

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$d)$ At [[6]]$^\\circ{C}$, the depth is $\\var{depth}km$ away from the surface.

", "type": "gapfill", "steps": [{"showCorrectAnswer": true, "showFeedbackIcon": true, "prompt": "

To find depth:

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To work out the depth at which temperature = $T^\\circ{C}$, you must firstly substitute in the value for $T^\\circ{C}$ and rearrange the equation given in the question statement to give an equation in the quadratic form $az^2+bz+c=0$.

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•  Hint: Subtract T from both sides of the equation.
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$z=\\frac{-b\\pm\\sqrt{b^2-(4ac)}}{2a}$, which will give you two values for depth, $z$.

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To find temperature:

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From the original equation given in the question statement, substitute the value given for depth, and solve for $T$.

To find depth:

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To work out the depth at which temperature = $T^\\circ{C}$, you must firstly substitute in the value for $T^\\circ{C}$, $\\var{t1}^\\circ{C}$, and rearrange the equation given in the question statement to give an equation in the quadratic form $az^2+bz+c=0$.

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$(-8.225\\times10^{-5})z^2+1.05z+\\var{constant1}=0$

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$z=\\frac{-b\\pm\\sqrt{b^2-(4ac)}}{2a}$, which will give you two values for depth, $z$.

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$z=\\frac{-1.05\\pm\\sqrt{1.05^2-(4\\times({-8.225}\\times10^{-5})\\times(\\var{constant1}))}}{2\\times({-8.225}\\times10^{-5})}$, which will give you two values for depth, $z$.

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To find temperature:

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From the original equation given in the question statement, substitute the value given for depth, and solve for $T$.

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