To keep it short, I am an Engineering undergrad. As the days go by, I keep coming closer to accepting the fact that I enjoy - and am passionate about - physics more than I am about Engineering. I know it would probably be easier to get a job with an Engineering degree(assumption?), and that...
Thank you for the responses! I have come to the conclusion that I just had a very simplified, romanticized view of math and mathematicians. I now see that even exceptional mathematicians(PhD in topology...thats epic) are human too and forget stuff/dont enjoy their work sometimes. I am growing...
Thanks for the advice! So basically, i should just focus on whatever topic is at hand, until i eventually reach other next level ones? In calc right now we're doing integrals of trig functions, so I should just forget what i don't know and focus on what i should know for the class?
What does it mean to be "good at math"?
I have recently began preparing for college by picking up a calculus textbook and doing it from the beginning. Although I'm not particularly talented at math, I am a little above average when it comes to understanding physics and mathematics concepts. I...
They are constants i suppose, because they are the initial condition which doesn't change. But i don't understand what "treat them as constants" means... does it mean i could just ignore them?
Tried another way..
From L=aLot-aLoto+lo
Y intercept= aLoto+lo
Slope=aLo
Homework Statement
For small changes in temperature, the formula for the expansion of a metal rod under a change of temperature is:
L-Lo = aLo(t - to)
L= length at temp. t
Lo= initial length at temp. to
a= constant that depends on metal
A) express L as a linear function of t. Find the...
The acceleration would be g i believe, or g plus the normal force of the rollercoaster/track system
I GOT IT OMG
V^2/R=g
V=(gr)^(1/2)
Mgh=1/2mv^2
Gh=1/2(gr)
H=1/2R
Thank you everyone. I believe this is the correct solution.
Ok so Ac at the top is v^2/r
Solving for velocity mgh=1/2mv^2 you get (2gh)^(1/2)
Plug that into v^2/r you get 2gh/r=ac
Solve for r you get r=2gh/ac... did i do something backwards? I can make it r/2=ghac But i feel like that's wrong, unless you take out the gh because theyre constants in...
I assumed it would stop because the car is conncected to the track. But if it was not connected to the track it would fall straight down.
To stay in a circle it requires centripetal acceleration, v^2/r
V^2/r= ac, v^2/r(1/2)=2ac? Or v^2/1/2r=mgh Am i on the right track? No pun intended...
If the v at the top of the loop=0, then the rollercoaster car will just stop there, so the initial hill has to be higher. I understand that, but why is a specific number such as 1/2R needed? Even if the initial hill was a very small amount higher, wouldn't the car be able to move enough to...
Homework Statement
A rollercoaster has an initial hill that leads to a circular loop of radius R. (a) Show that the top of the hill must be at least 1/2 R higher than the highest part of the loop. (b) Discuss additional factors that must be considered in the design of an actual rollercoaster...