Not a man of them's worth a wooden nickel, but Lord above, how I do love the wretches. Mine are rogues, each and every one of them. Where are they?'Ī spell, is it? he asked her, but the question was really meant for me.That I can think of nothing but you, no matter where I go, or what I do? That in each victim I take, I find you? Oh, yes, think on it, Merrick, think on what I do to survive, don't please live in dreams. Poirot remained lost in thought for a few minutes. So we have 1512 divided by 5.6 times 10 to the minus six.There was silence. By the way, I So we have the Coombs that charge Q from Part A and were given the average current so we can plug in our numbers.

And we want to find find out How long can this pacemaker last with that average current? So again, the equation for current is cute amount of charge divided by time. The amount of charge we get 1502 mhm columns of charge and in part B were given the average current that this pacemaker has, which is given as 5.6 times 10 to the minus six camps. Canceled hours cancel leaving coolers by itself. We have 0.42 ham powers and again and these columns for second times our So we use those conversions. So we will have cue from part A, and we'll have the current that's given to us in part beam and then we simply solve for T Southern Part A. So then, in part B, to find how long this pacemaker last we use the equation for current, which is current equals Q two columns of charge divided by t.

And then once you multiply the conversions, we end up getting how much columns of charge the pacemaker has. So the second's cancel the hours cancel minutes, and we're just left with columns. So I want to have columns isolated, so use a few conversions so you can say one hour, 60 minutes, and one minute is 16 seconds. So what we have here is actually columns per second times an hour. We know the amps is actually coolants per second. To find the columns of charge, we need to use the unit of amps. How long can this pacemaker last? So in part A. You want to find out how many columns of charge it has been? A part B.

So this is just telling you that it has a 0.42 amps of current per hour, and in part A.

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