Recent content by BitterSuites

Oh no. *tries not to be obviously deficient* Ok, A has g pulling down and the string pulling up. B has g pulling down, the string pulling up, and um...something else having to do with the spin? o.o Torque maybe? (Please don't make fun of me :D)

Oh, thank you so much *sighs in relief* Converting correctly makes the problem work. Who knew? :D

Uh oh. Did I go the wrong way with it? 43000N/m?

That would be gravity, correct? Disk A (the one tethered to the string) would have g pulling down and B pulling it up. Disk B (the yo-yo) would have g pulling down, A pulling it up, and centrifugal force (though it isn't really a force, so maybe I'm wrong) Am I stumbling the right direction?

Thanks for the catch on conversion. I changed length to .88 and K to 4.3, getting an answer in m. I then converted it to cm but the online system still says I am incorrect. I can't use mv^2/r because I don't have v. That is why I did that insane derivation. Or am I wrong? I think I'm a few...

Homework Statement Two uniform disks are connected by a light inextensible string supported by a massless pulley on a frictionless axis. (Very ideal LOL) The string is attached to a point on th circumference of disk A and wound around the disk B like a yo-yo. Their moment of inertia...

[SOLVED] Unstretched Spring and Angular Velocity Problem Homework Statement A disk of mass 73 kg is held by a spring of unstretched length 88 cm and stiffness constant 430 N/cm, and rests on a frictionless table. It is then orbited in a circle at angular velocity .675 rad/s. By how...

So F = m(a+g) = 523 (.469657 + 9.8) = 5371.03 d = .89 * 3.79 = 3.3731 W = Fd = 5371.03 * 3.3731 = 18117W becomes 18.117 kW Where am I still going wrong? This was not correct.

Ok. So a = 1.78/3.79 = .469657 avg v = (V + Vo)/2 = (1.78 + 0)/2 = .89 So, d = avg v * t = .89 * 3.79 Am I at least on the right side of the highway?

Hmm. I guess it is missing d, as in W=Fd. Am I capable of calculating d?

Homework Statement A 523 kg elevator starts from rest. It moves upward for 3.79s with a constant acceleration until it reaches its cruising speed of 1.78 m/s. The acceleration of gravity is 9.8 m/s^2. Find the average power delivered by the elevator motor during the period of this...

Homework Statement A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 7.51m. The acceleration of gravity is 9.8 m/s^2. Given: d = 7.51 theta = 32 m = 10.1 g = 9.8 coefficient of friction = .178 v = 1.77 1...

*sigh* I keep making really silly mistakes. So I input the correct equation but it came out correct. v=2pir/t 7723.55 m/s = 2pi * 6689000m/T 7723.55T = 4.20282e7 T = 5221.57 seconds Convert to hours 5221.57/60 = 90.6928 minutes 90.6928/60 = 1.51155 hours Thank you so much.

[SOLVED] Centripetal Acceleration Part II Homework Statement (From previous question) In order for a satellite to move in a stable circular orbit of radius 6689 km at a constant speed, its centripetal acceleration must be inversely proportional to the square of the radius of the orbit...

Ah :) Thanks for everyone's help! I'm marking it solved now.