Hi guys. Ok I'm working on this problem, it's more theoretical than anything.:yuck:
I'll try to explain as best I can. Lets say you have a runner going past the following points up till and past 30 m:
Distance (x) of x1=0, x2=4, x3=9, x4=20, and x5=30m
the times to these intervals are 0...
I see. So V=d/t, then use that as a final V in a=Vf-Vi/t, a=.4 m/s^2 on a non frictional surface.
Sorry, I just didn't get the formulas Xiankai was using.
I see! so if I had a coefficient of friction then I all I would do is go (40kg x 0.4 m/s^2) - u (mu) x F normal (m x g)...
Ok,
So, lets say I was to drag a 40kg load in a set distance of 10 meter. I record the time, it was 5 secs.
Now I find power, P=work/t. We know work=force x distance.
SO my question is, is force just mass of load x acceleration due to gravity? 9.81m/s^2.
So P= ((40kg x 9.81) * 10...
Wait, got an idea...What about if he rotates his leg in the opposite direction of his hand rotation?
Ok..I'm going to go bang my head on the table now lol. This is nuts.
See, now I'm really confused...Ok, it says describe two ways which we can stop his twisting while still in the air.
So what Lightgrav suggested was 1)lift his right hand out of page and drop left one--wouldn't this cause him to start twisting the other way?
2)aren't his feet always beneath...
So are you saying that if my guy spreads himself out, he'll eventually stop twisting because of his large Inertia which gives a small angular velocity as in H(ang momentum)=Iw? So can I say he can spread his hands out and spread his legs out as two different movements OR are they the same, ie...
So, in my diagram I outlined the directions, did I do these right?
I don't think he can "stop" his movement of falling down, but he can stop twisting. As I've noted, he might be able to life his right hand OR drop his left but are those considered different? By doing that he balances out his...
Hey:smile: ,
The question:
The athlete on the diagram (linked) is free in the air and has a total angular momentum as indiated by the red vector.
1)Indicate with a vector the direction of his twist based on his orientation and angular momentum, show how you got this.
2)Describe 2...
The other case...
Calculate the average force required to absorb the energy of your landing if you take 0.25 meters to stop.
F=m(v2-v1)/t
=68(0-3.05m/s)/t
=68(3.05m/s)/0.082
=2529N?
t=d/v
t=0.25/3.05
t=0.082
Ok, I think I got it?
on the question where it asks:
Calculate the average force required to absorb the energy of your landing if you take 0.25 meters to stop.
Could it really be this simple?
I know my energy for the jump is 323.5J. And to apply such a force to stop at 0.25m, I'd...
You're lots of help! I wonder...I posted this https://www.physicsforums.com/showthread.php?t=96722 maybe you could give me some input. Much appreciated :).
Ahhh, my sticky point was the distance from my backpack to my toes. Thought I had to calculate that :rofl: :rofl: . Ok, I think I got this. Thanks for the help.
Ok...so from your formula you're saying that:
Mass total (ie me and the backpack) x the length of C.o.m. which I already found=sum of (M_i? x l_i?> I don't get this part. what is i?
Also, I don't know the value for the distance from the center of mass of backpack to the fulcrum or my...
Hi,
Ok so we were told to find our center of mass using the reaction board method, but modified. We just do a pushup position with our hands on a scale, I have a electronic one and toes on the floor.
My values obitained are:
weight=68kg
the scale reading in a pushup position=48.9kg...
I really don't get what this 0.25 meters come into play...I asked my teacher and he keeps telling me, use what info you got and find what you don't know, that's the key. LOL, uh yeah that's great.
So, wait wait, in the 180 position, the quad is still working upwards right? not horizontal because then it would be acting through the axis of rotation, which gives you no torque force.
I get what you're saying, it says moment arm should be constant lol, my bad. But are you also saying that...
The question is:
You are wearing a special 10 kg weight boot on each of your feet and you're sitting on a table with your legs hanging over the edge at the knee joint. The knees are held at an angle of 180 degrees (straight out) and 135 degrees. In each case how much force must be produced by...
I know, it doesn't make any sense, but that's exactly what it says on the worksheet.
I also don't get the wording on this one..
Calculate the average force required to jump this high if you apply the force over a distance of 0.5 meters. ???
Thanks.
Ok :rofl: ,
I had to test my maximum vertical jump. I did three trials and here's what I got:
-Maximum standing height (ie without jumping reaching with one hand) = 221.5 cm
-Maximum jumping height with one hand (best of three) = 270 cm
-Net Height 270-221.5 cm = 48.5 cm, 0.485m
My weight...
Ok so I carried out the experiment: :cool:
Runners Dress Shoes
Trial Height of board before slip (H) Height of board before slip (H)
1 59.9 48.5
2 57.6 48
3 58.5...