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DarkBlueChocobo
 one year ago
Help with application problem
DarkBlueChocobo
 one year ago
Help with application problem

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DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0A paddle boat is traveling down a lake at 15 mph. If the paddle wheel is 8 ft in diameter, how many revolutions per min is it rotating

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443836986859:dw

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0so the radius of the wheel is 4 ft then

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0but first wouldnt we have to change units

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443837060765:dw

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443837124936:dw

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0Doesn't this give us the angular speed?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Umm umm umm, I'm getting a little confused by your break down of the units. Lemme show you how I'm thinking about it.\[\large\rm boat~speed=\frac{15\color{orangered}{(miles)}}{1\color{royalblue}{hour}}=\frac{15\color{orangered}{(5280ft)}}{\color{royalblue}{60\min}}\]So we get that the boat travels at a rate of:\[\large\rm boat~speed=\frac{15(5280)}{60}\frac{ft}{\min}\]And the circumference of our wheel is:\[\large\rm C=\pi d\qquad\to\qquad C=8 \pi ft\]The distance around the wheel, the circumference, is `one full revolution`. So our wheel is moving at this rate:\[\large\rm wheel~speed=8\pi\frac{ft}{rev}\] So then ummmmm, we divide I guess? \[\large\rm \frac{15(5280)}{60}\frac{ft}{\min}\div8\pi\frac{ft}{rev}\]Sorry I'm a little rusty on these types of problems :) lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm =\frac{15(5280)}{60}\frac{\cancel{ft}}{\min}\times\frac{1}{8\pi}\frac{rev}{\cancel{ft}}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Mmm I need a brief refresher on this stuff XD What is this type of problem called? I'll Google it. Something with rotational motion?

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0Well the beginning of it was Angular speed and then you converted it to linear speed basically because they are related

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I guess I made it a little bit of extra work by writing the angular relation like this. I should have written it like this: \(\large\rm \dfrac{1rev}{8\pi ft}\) And then it's just a multiplication problem, canceling out the units in the long string as you did.\[\large\rm \left(\frac{15miles}{hr}\right)\cdot\left(\frac{5280ft}{mile}\right)\cdot\left(\frac{hr}{60\min}\right)\cdot\left(\frac{1rev}{8\pi ft}\right)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I think you just had your last term a little off, ya? :o The wheel spins 8pi every revolution.

DarkBlueChocobo
 one year ago
Best ResponseYou've already chosen the best response.0It seems so :<
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