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ok so lim(x->1+)=(1+1)=2 lim(x->1-)=(1+1)=2 so the limit exists, i don't understand how that helps in finding the overall limit of F(x) at 1.

sorry that should be x+1

F(x)= 2x^2, x<1 3, X=1 X+, x>1 Find lim(x-->1) f(X) f(1)= not exactly sure how to do this. would not f(1) be just 3, since it is defined as that in the function. I am not sure about how to take that limit. Thanks

lim(x->pi/4) tan(2x) =tan(pi/2) wat to do after this, it is undefined

what is the difference between amino acids on their own and amino acids that have been incorporated into a polypeptide chain? My answer is that amino acids on their on do nothing. However when they interact they create a protein which eventually takes on a particular function. why does...

Hi Doc Al, Almost solved it the a i use here will be angular acceleration so net torque=T*R or Ia=T*R Ia=(mg-mRa)R Ia=mgr-mR^2a Ia+mgR^2a=mgr a(Ia+mR^2)=mgr a=mgr/I+mR^2 for acceleration the tension is a=mR^2g/(I+mr^2)

An object of mass m is tied to a light string wound around a wheel that has a moment of inertia I and radius R. The wheel bearing is frictionless and the string does not slip. find the tension and the acceleration of the object. I think mg-T=ma and that I=MR^2 thats all i know can...

ok i understand. What i don't understand is how to write a "physics story". I guess that i have to describe what's going on in the problem?

Chubby Santa(mass 100kg) decides to have some fun and slides 9.0 m down a snowy roof)starting from rest. There is a 135 n fricitional force between santa and the roof. After sliding the 9.0 meters he collides with and clings to, an elf (mass 50kg) who was sitting on the roof edge. They both fall...

yah i got the answer as -.57m/s. thnx

then i would get a negative v since v'=325-450/40=-3.125m/s

so the m_{2}v_{2) becomes negative? i don't understand what u mean by "direction matters"

A 130-KG tackler moving at 2.5 m/s meets head on(and tackles) a 90 k-g halfback moving at 5.0 m/s. What will be their mutual speed immediatley after the collison? someone please check my work m_{1}v_{1}+m_{2}v_{2}=v'(m_{1}+m_{2}) so v'=\frac{325+450}{220} v'=3.52 m/s

Hey Doc Al, thnx for helping me through this process :). I got \theta_{2}=-33 v'_{B}=.81 m/s

yes i plugged in numbers and i got .068=v'_{B}\cos\theta'_{2 -.72=v'_{B}\sin\theta'_{2} thats all i can do to simplify

can someone help me in simplifying these equations? thnx

well I am not sure whether this problem has a definite answer(as in 1.2 or 3.4 etc)or whther I am suposed to solve in terms of variables

A billiard ball of mass m_{A}=0.400kg moving with a speed v_{A} =1.8m/s strikes a second ball, initially at rest, of mass M_{B}=0.500kg. As a result of the collision, the first ball is deflected off at an angle of 30\deg with a speed of v'_{A}=1.1 m/s. a) taking the x-axis as the positive...

So the kinetic energy would be 1/2*184(0.052)^2=.25J

so the total energy would be 1/2 KA^2

thnx a lot man

yes but i get v=0 when i do the derivative. is this correct? thnx for ur help

ok so i have to take the derivative x=3.0\cos(5\pi t+\pi) so the derivative of that would be 3.0*-\sin(5\pi(0.5)+\pi)*d/dx(5\pi(0.5)+\pi

yes i need to calculuate velocity, but i don't know how to get it.

I have no idea what you are talking about? are my other answers correct?

A particle has displacement given byx=3.0\cos(5\pi t+\pi) where x is in meters and t in seconds A)What are the frequency ,F, and the period,T, of motion B)What is the greatest distance the particle travels from equlibrium c)Where is the particle at time t=0?t=0.5(s) D)What is the velocity...

At t=0, a 650-g(0.65kg) mass at rest on the end of a horizontal spring(k=184n/m) is struck by a hammer which gives it an initial speed of 2.26 m/s. Determine (a) the period and frequency of the motion (b)the amplitude (c)the maximum acceleration (d) the position as a function of time (e) the...

hey everyone i need help with this antiderivative \int\sec^{2}x\tan^{4}x my guess is that it is tan^5(x)/5

ok i got v=1.59 for part A and v=1.18 for part b how do i find the total energy of the system?

ok i found K as 1.27 and plugged it back so \ so i got the velocity as 0.28m/s is this right so then the velocity from 0.1 m from equibium would be .78m/s c) total energy of the system would be 1/2KA^2=0.014--dosent make sense thnx

spring mass help--desperate A 0.35-kg mass at the end of the vibrates 3 times per second with an amplitude of 0.15m. Determine (a)the velocity when it passes the equilibrum point (b)the velocity when it is 0.10m from the equilibrum point(c)the total energy of the system(d)the equation...

yah I found the answer as x=ln[1+sqt2]----thanks for your help

at what point on the curve y=\cosh x does the tangent have slope 1 I have no idea how to approach this problem my work 1=\sinh x\frac{dy}{dx} \frac{1}{sinh x}=\frac{dy}{dx}

I need help proving this hyperbolic function Prove that \tan^{-1}\hbar {x}=\frac{1}{2}\ln\frac{1+x}{1-x} my work x=e^y-e^-y/e^y+e^-y (e^y+e^-y)x=e^y-e^-y 0=e^y-e^-y-xe^y+xe^-y e^y(e^y-e^-y-xe^y+xe^-y) e^2y-x(e^2y)-1+x=0 I know i have to use the quadratic equation here

Hey guys I need help on this one physics problem. I did the problem but I am not sure whether I did it right. Please make comments if i did the problem wrong. Thank you. A projectile is shot from the edge of a cliff 125 M above ground levbel with an inital speed of 65 m/s at an angle of 37...