Recent content by rent981

Yes that was a typo. So...\sum (from 0-infinty) (-3/5)(4/5X)^3n would be ok? Sorry this is a new topic for me. Thanks you so much for helping!

does that look ok?

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n (-3/5)(4x)^3n

oops. I am meant to write [infinity]\sum[/n=0] (-3/5)(4x)^3n.

Ok if I rewrite I need to make the first term of the denominator 1. So I divide by -5. So then does the sum from 0 to infinity of (-3/5)(x)^3n look better?

[b]1. Find a power series for F(x)= 3/4x^3-5, where c=1 [b]2. power series = 1/a-r [b]3. What I did was take a derivative to get a similar function that was easier to solve. I used 1/x^3-1. Then I found a series for that function. Which I got \sum(x^3)^n. Then I added back from my...

Here is the problem I am working on. Here is the work I have so far, does this look right. A grinding wheel that has a mass of 65.0 kg and a radius of 0.500 m is rotating at 75.0 rad/s. (The carousel can be modeled as a disk and assume it rotates without friction on its axis) a) What is...

so if friction is ignored, in a game of tug of war, the opposing sides of the rope would meet at the center of mass? Provided there are no other contributing factors.

well actually 1m should be zero, but it still works out to be 500.

The center of mass would be (100kg)(1m)+(500kg)(600m)/(100+500kg)=500.167 m

hehe nice. Thanks for your help. I understand this concept a lot better now!

Well if the forces are pointing in the same direction the sign of work would be positive. Because the equation would reduce to w=Fs.

[b]1. Well this might not be exactly a tug of war problem but its pretty close. A 100 kg fisherman and a 500 kg supply crate are on a frozen pond that is virtually frictionless. The man and the crate are initially 600 m apart. The fisherman uses a very light rope to pull the crate closer to...

The question states what is the sign of the work done by gravity not the person, sorry for the confusion.

[b]1. If your are walking down a flight of stairs would the sign of work be negative, positive or zero? [b]2. Well work =FCOS Theta Delta X. [b]3. SO I think it would be negative. My reasoning is that the normal force of the person walking down a flight of stairs is in the opposite...