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CDQuizAns - 7B Quiz 1 Last 6 digits of ID Name Signature DL...

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7B Quiz 1 Name: DL Sec. Last 6 digits of ID # Signature: Physics 7B Quiz 1 Spring Session 2008 Larger human arteries (left figure below) are typically elastic, so the artery walls are supported against the pressure of the surrounding tissue by the blood within them. In FNT 5.1-6 you considered an aneurysm, an abnormal enlargement of a section of an artery. A cholesterol plaque is a deposit which can form in an artery and result in an abnormal narrowing of a section of artery (second figure from left). Two surgical treatments are possible. The first is to place a stent , a thin wire mesh tube, inside the artery to hold it open (second figure from right). The second is a bypass , in which another artery is placed parallel to the narrowed section of original artery (right figure). 1) The stent is used to keep the artery from collapsing under the pressure of the surrounding tissue. This is not a danger in a healthy artery. Why is it a problem when a cholesterol plaque has narrowed the artery? Use the models we have learned about in class to explain this. Which term(s) in the equations we have been using are involved in your explanation, and how? 2) Now consider the treatment by the use of a bypass. How does this treatment work to improve the supply of blood to the heart? Use the models we have learned about in class to explain this. Which term(s) in the equations we have been using are involved in your explanation, and how? Answer: Note that the bypass increases the total cross section from point 3 to point 4. This reduces the speed of the blood through the diseased region in general, with the result that the danger of collapsing the region with the plaque is reduced.
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7B Quiz 2 Name: Answer Key … Kevin Klapstein DL Sec. Last 6 digits of ID # Signature: Physics 7B Quiz 2 Spring Session 2008 No books or notes. Calculators OK. Note that a few equations have been included at the bottom of the quiz. Show all your work below. Answers alone do not receive credit! In the diagram at right, the current through R 1 is I 1 and the voltage difference across R 1 is D V 1 , and similarly for R 2 , R 3 , and R 4 . The values of the resistors are: R 1 =20 W , R 2 =20 W , R 3 =50 W , and R 4 =75 W . The battery produces an emf of d =80Volts. The symbol attached to point D is a symbol for “ground”, implying a voltage of 0V. FOR ALL PARTS OF ALL QUESTIONS, YOU MUST EXPLAIN YOUR REASONING. 1) The right side of the battery (point D) is at 0 volts. What are the voltages: a. At point A, on the left side of R 1 ? Explain your answer. Answer: V A =80Volts. From D at V D =0V to the rhs of the battery the potential increases to d =80Volts. This point is connected to point A by a wire, so that the voltage remains constant. b. At point C, on the right side of R 3 ? Explain your answer. Answer: Point C is connected by a conducting wire to point D, which is grounded, so V C =V D =0Volts.
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CDQuizAns - 7B Quiz 1 Last 6 digits of ID Name Signature DL...

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