# HW1 - 1.6(Smoots(Willies(Zeldas How many Willies and Zeldas...

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From the figure we see that: 212 S = 258 W giving the conversion factor 258 W 212 S Then, 258 W 50.0 S 50.0 S 60.8 W 212 S ⎛⎞ == ⎜⎟ ⎝⎠ To find the number of Zelda’s note that the origin of each axis (Zeldas and Smoots) is not the same. Now to find the conversion we must use (212 -32) S = (216-60) Z 156 Z 180 S Then, 156 Z 50.0 S 50.0 S 43.3 Z 180 S 1.6 How many Willies and Zeldas in 50.0 Smoots? (Smoots) (Willies) (Zeldas) Note that we could have formed the inverse ratio for the conversion factor but only this one results in the Smoots canceling to give Willies. (in words 258 Willies per 212 Smoots)

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1.14 The pulsar rotates once every 1.55780644887275 ms with an error of ±3 in the last digit. a) How many rotations does the pulsar make in 7.00 days? The pulsar rotates with a rate of 1 rotation per 1.557x10 -3 seconds or: Where we converted the milliseconds to seconds (1 ms = 10 -3 s) and we need carry only one more digit in the calculation than the parameter having the fewest significant figures to be used in the problem. In 7.00 days the pulsar then rotates: 3 1rotation rotations Rate 642.3 1.557x10 s s == 8 hr min s rotations (7.00 day) 24 60 60 642.3 3.88x10 rotations day hr min s ⎛⎞ = ⎜⎟ ⎝⎠ to 3 sig. figs. (or 3.88x10 8 times)
b) How much time does it take the pulsar to rotate exactly 1 million times? 6 63 3 10 rotations (10 )(1.55780644887275x10 s) 1557.80644887275 s 1rotation 1.55780644887275x10 s == c) What is the associated uncertainty? The uncertainty in the time of one rotation is ± 3x10 -17 s (count the number of decimal places to the last digit of the rotation time, which, as the problem states, is what the ±3 refers to, and remember that it is in milliseconds). Hence the uncertainty in the time taken for 10 6 rotations is (10 6 )(± 3x10 -17 s) = ± 3x10 -11 s Note that you can not get this many significant figures on your calculator. Here we used the decimal point shifting properties of exponential notation.

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1.30 0.8 m(t) 5.00t 3t 20 =− + With m in grams and t 0 in seconds. a) At what time is the water mass greatest? To find when the function is an extremum, differentiate m(t) wrt t, equate to zero and solve for t: 0.8 1 dm 5.00(0.8t ) 3 0 dt = −= 0.2 3 t 5.00(0.8) = 0.2 1 0.75 t = 0.2 1 t1 . 3 3 0.75 == 1 0.2 t 1.33 4.21s Which is the time at which the mass is maximum (see plot). Here’s a plot of this function (you could do this in Excel or other graphing capable software) 05 1 0 20 22 24 23.16 20 mt () 10 0 t The graph is not necessary for solving the problem but it often helps. Here for example rather than needing to differentiate again to test if we have a minimum or maximum, the plot shows it to be a maximum.
b) What is the greatest mass? Insert into the original m(t) the time just found 0.8 m(4.21s) 5.00(4.21s) 3(4.21s) 20 =− + max m 23.2 gm = c) What is the rate of mass change (in kg/min) at time t = 2.00 s ? The rate of mass change is given by the derivative with time, 0.8 1 dm 5.00(0.8t ) 3 dt At t = 2.00 s this is 0.2 dm gm (t 2.00 s) 5.00(0.8(2.00) ) 3 4.82 dt s == = 2 gm kg s kg 0.482 0.001 60 2.89x10 sg m m i n m i n ⎛⎞ ⎜⎟ ⎝⎠ Convert to kg/min:

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d) What is the rate of mass change (in kg/min) at time t = 5.00 s ?
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HW1 - 1.6(Smoots(Willies(Zeldas How many Willies and Zeldas...

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