Recent content by aleferesco

Thanks a lot!, I got the answer

I'm sorry I don't understand the comparison. 1/2k(x[SIZE="1"]2)^2 = 1/2k(2x[SIZE="1"]1)^2 1/2k(7)^2 = 1/2k [2(7)^2] 24.5k = 98k After this I have no idea what to do... am I supposed to have it like this W[SIZE="1"]2 = 98k/24.5k (the k gets canceled out) W[SIZE="1"]2 = 4...

Homework Statement In order to compress a spring by 7cm from its natural length, 20J of work has to be done. How much additional work should be performed in order to compress the spring by an additional 7cm? Homework Equations F= -Kx (spring's equation) W=1/2(Kx)x = [k(x)^2]/2...

Homework Statement A basketball is dropped from a height of 2m. It lost lost 20% of its energy, in the bouncing from the floor due to friction in the air, before it reaches the highest point of its first bouncing. How high did it bounce? h1= 2m h2=? g=9.8m/s^2 Homework Equations...

(-∞,0) u [0,5] u (5,+∞) ? do I always have to include for example the [0,5] when function is XER? thanks

I just have a question which I'm confused about, this is the same right? Domain is (-∞,0) U (0,+∞) and Domain D: {XER, x ≠ 0} and this is the same right? Domain is (-∞,0) U (5,+∞) and Domain D: {XER, x ≠ 0, x≠ 5} Attempt : I know that U is used to...

is glucose soluble in diethyl ether? my guess is that glucose is not soluble in ethers, but I don't know why, can somebody explain this to me please? thanks

Thank you very much!

so omega= 2.35s^-1 and so to find frequency I could just do f=2.35s^-1/2pi I'm not sure about the reversing 1/1/T = s^-1

Homework Statement The position of a mass oscillating on a spring is given by x(t) = (18.3cm)cos[(2.35s^-1)t] . What is the frequency of this motion? Homework Equations X= Amplitude x Cos (2pi/T x time) The Attempt at a Solution I know that frequency= 1/Period, I am trying to...

1) If the limit of the function g(x) = [f(x) -8]/[x-1] is equals to 10 as x approches 1, then find the limit of f(x) as x approaches 1 2) If the limit of the function h(x) = f(x)/x2 is equals to 5 as x approaches 0, then find the following limits a) limit of f(x) as x approaches...

but when would you say you feel lighter, at the beginning, middle or end of the elevator ride??

I'm just wondering, when I'm in an elevator going up/down when do I feel lighter?

ooo! 0 = at2 - 2v0t - 2(s0)c t = (2v0t +- [(-2v0)2 - 4(a)(-2(s0)c)]1/2/2a since time is only positive then... I only take the positive side of the obtained values.. tmax = (2v0t + [4(v0)2 - 8(a)(s0)c)]1/2/2a

so if what I did was correct then now I have to Isolate for t, at2 = 2v0t + 2(s0)c t2 = (2v0t + 2(s0)c)/a t2 = (2(v0t + (s0)c))/a t = [(2(v0t + (s0)c))/a]1/2 but my question is, what can I do with the 't' that is inside of the equation does it seem right?