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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
a car has a mass of 1500kg, and is traveling at 26.8 m/s at the top of an incline; the incline has a slope of 1/10 a) what is the Kinetic Energy? i did 1/2 m v^2 b) if the car coasts downhill, how fast is it going after it has descended 1000 meters? i tried: \(v^2=(v_o)^2+2a\Delta x\) and tried to determine the acceleration. c) What is the breaking force needed to maintain the initial velocity going down the hill? i considered: F = ma, using the acceleration from part b) d) at the bottom of the hill it goes around a curve at 29 m/s; the curve has a radius of 90 meters, and the tires have a coeff of kinetic friction of .98 Does the car slide, or does it stay on the curve? My idea was: Fk = uk m g , Fc = m ac, where ac=v^2/r
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354643097910:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[a=\frac{F}{m}=\frac{mg~\sin\alpha}{m}=9.8(.1)=.98\frac{m}{s^2}\]
 one year ago

MuH4hA Group TitleBest ResponseYou've already chosen the best response.2
If your idea for d) is, to compare Fk with Fc and then draw the conclusion, I'd say everything looks fine. One quick thing would be: You'd make it a lot easier to read, if you'd use the built in latexthingy ($$) and elaborated what the different symbols are supposed to be. Makes spotting mistakes a lot simpler ;)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
``` i use \(\) for inline, and \[\] for new line ``` i could never get the hand of the $$s :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
but as long as it looks fine, i do feel better about it :) thnx
 one year ago

MuH4hA Group TitleBest ResponseYou've already chosen the best response.2
Well, if you want to become a scientist, you're gonna have to learn latex anyways, I'd say.. so why not start now? Here[1] is a great introduction, if you feel like it. Also, things like \[v = \pm \sqrt{v_0 ^2 + 2\,a\,\Delta x }\] aren't that obvious. (Maybe they are and I am just too dumb to see it ^_^). So if you'd include the couple if lines how you got there, maybe you get a faster answer. 'Cause I could see one thinking: I won't bother getting out a sheet of paper and trying to figure out if that's ok or not. Whereas if you just write down _some_ steps, I can just read along and check in my head ;) 1. http://tobi.oetiker.ch/lshort/lshort.pdf
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
that was one of the 3 equations for motion in the textbook, its useful for things that dont have an explicit time component
 one year ago

MuH4hA Group TitleBest ResponseYou've already chosen the best response.2
Oh ok.. Well, if you are allowed to "just use" it, without deriving it first (yeah in this case, it's pretty simple) that's fine. I'm just saying: In general, the guy trying to help you, won't have all this equations in his head, nor will he have the same textbook you got  especially not opened on the same page in front of him ;) No big deal in this case, just... when things get a little more complicated, you should elaborate. And of course: you shouldn't just accept that equation as given anyways  ever. Aside from the possibility of a misprint, physics is about understanding and being able to get there without memorizing formulas for every possible situation. Anywhooo.. nice job on the problem, keep it up ;)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
got it, thnx :)
 one year ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
A couple of notes: a) Good. Pretty straight forward here. b) Is the object descending 1000 meters in the vertical direction, i.e. does it's altitude on a barometer decrease by 1000 meters. Or does it travel 1000 meters along the inclined plane? It looks like you assumed the later. In the case that it is the former, it is convenient to use energy. \[\Delta KE = \Delta PE \] c) Looks good. A balance of forces creates for steadystate. d) I hope your tires don't have kinetic coefficient of friction! It means they are locked up. Remember that tires roll without slipping, meaning the contact patch isn't moving relative to the ground. This is why ABS is awesome. Keeps us the static coefficient area, which is usually greater than kinetic coefficients. Other than that, your approach is fine. As long as the frictional force of the tires is greater than the centripetal force, you're staying on the road.
 one year ago
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