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A wealthy uncle wishes to x an endowment for his favorite nephew. He wants the fund to pay the young fellow $1,000,000 in cash on the day of his thirtieth birthday. The endowment is set up on the day of the nephew s birth and is locked in at 8% interest compounded continuously. How much principle should be put into the account to yield the necessary payoff (a) (b) (c) (d) (e) 88,553.04 90,717.95 92,769.23 91,445.12 90,551.98
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Figure 5-8
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somehow has a very large kinetic energy, then it may be able to penetrate into the bilayer, at which point moving out of the bilayer to either side is highly favorable.) Large molecules are also unable to pass through the bilayer, unless they have enough energy to disrupt the forces that hold the bilayer together. Being large, they have to push aside more lipid molecules in order to pass through than a smaller molecule would have to do. So, they also have to have enough energy to disrupt more of the forces holding the lipid molecules together. On the other hand, small, uncharged, nonpolar (or only slightly polar) molecules can pass through the bilayer with relative ease. The forces that hold the bilayer together are dispersion forces, the hydrophobic effect, and, if cations are present, an additional stabilization by the cations acting as counterions to reduce the repulsive force between the phosphate head groups. Obviously since cations stabilize the bilayer, their presence has a strengthening effect on the bilayer. This makes it more difficult for large molecules to enter, since large molecules have to somehow push the lipids aside. Pushing the lipids aside requires disrupting the dispersion forces that attract the hydrocarbon tails to one another. This takes energy, and the larger the molecule, the more lipid molecules it has to push aside (and the more energy required). On the other hand, at lower concentrations of cations, the repulsive force of the head groups may assist the large molecules getting through, by making it easier to push the lipids aside. The strength of the dispersion forces depends largely on the amount of close contact between the hydrocarbon chains. This in turn depends on how close together the chains can pack and how long the chains are. Longer chains have more contact over which the dispersion forces can attract adjacent molecules. So, longer chains make stronger dispersion forces and more stable bilayers. (Longer chains also mean a thicker bilayer and thus a greater distance over which a molecule passing through the bilayer has to travel.) Another significant factor is whether the hydrocarbon chains are saturated, unsaturated, or polyunsaturated. Unsaturated chains (recall from Chap. 7) have one or more double bonds between some of the carbon atoms. Double bonds restrict free rotation, so each chain will be stiff at the location of the double bond, with a particular angle or kink in the chain at that location. The end result, due to these kinks, is that unsaturated and polyunsaturated chains prevent the lipid molecules from packing as closely together as saturated lipids can. This is illustrated in Fig. 11-6. Dispersion forces are proportional to 1/r 6; that is, they decrease in proportion to the sixth power of the distance. Less tightly packed chains mean that on average the chains are further apart from each other; this leads to weaker dispersion forces.
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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Fig. 3-15 The Thevenin equivalent voltage for Example 3-4.
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Click here to open Edit Line Style dialog
shB shF Idle shl shA0 shD Stage PTQ Rate monitors
Analyze and Conclude
Optionally you can globally change the timeout values for your data SAs. (You can also override the global timeout on a peer-by-peer basis in a crypto map entry, discussed in the next section.) Once these timeouts are reached, the appliance will either rekey them or tear them down and rebuild the SAs. Here s the syntax to change the global SA timers:
This will provide the larger required capacitance value dictated by the C1 and C2 capacitors, as calculated for Fig. 6.57. However, the relative tuning range will obviously decrease as the design frequency is decreased, since CV will now have far less of an effect on altering the increased fixed capacitance of CS (Fig. 6.58). In addition, the minimum value of capacitance will be set by the value of CS. Nonetheless, more varactors can be paralleled to increase the tuning range, as required, and more varactor-tuned LC tanks can be added to increase the selectivity of this tunable filter.
Dolby Digital (AC-3) Audio Coding
SOLUTION We observe that f (x) = sin x 0 for 2 x and 0 x . Likewise, f (x) = sin x 0 for x 0 and x 2 . As a result Area =
EXAMPLE
1.16 sec
22:
Application
Imagine that you are using array initialization to build a table of error messages, as shown here:
Established Command Configuration
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