Recent content by Wobble

It's supposed to be in Kelvin? I feel dumb now. I got them both. Thank you for your help.

Homework Statement The engine of a Ferrari F355 F1 sports car takes in air at 20.0 *C and 1.00 atm and compresses it adiabatically to 0.0900 times the original volume. The air may be treated as an ideal gas with \gamma = 1.40. Find Final Temp and Final Pressure Homework Equations...

Looking a wikipedia 1. (2piR)/V=T 2. (2piV)/a=T 3. a=(v^2)/R and for it in radians per second 4. omega=2pi/T So do I just want equation 3, substitute it into F=ma, and get (M+m)(v^2/R)=answer for C? It says "frequency of the vibration" though, which doesn't sound like a force answer...

Torque= R cross F= I times alpha Alpha is then a/R. But I need a velocity in there somehow to solve part C. 50mph isn't an angular velocity, nor does a/R contain velocity in it. How do I account for the change in the center of mass? Its going to involve my answer to A somehow, which I got to...

Well, I could say that the perfect tire's R for the center of mass equation is 0, and that would just give me 0 as the center of mass. Then the center of mass for the two objects would be: (M*0+m*R)/(M+m) and that would give me my new center of mass, which would be the distance between the...

Well for the center of mass of the tire, I don't really have a radius for it? The mass m is 1 ft away from the center of the tire, but how far away is mass M from the center of the tire? Do I use 1 ft for it a well? But then I would get (M+m)/(M+m) which is just 1... (\Sigmamr)/(\Sigmam) is...

Homework Statement Homework Equations \Sigma\tau= r x F = I*\alpha Angular Momentum L= r x p = r x mv = r x mr\omega = mr2 * \omega2 editediteditediteditediteditediteditediteditediteditediteditedit This formatting looks weird. It should be r^2 *(omega)^2 and I*(alpha for the next part)...

Ok cool. I said something different earlier, but you went with the correct numbers.

I found an error. The mass of the asteroid times rocket velocity is 5*10^13, NOT 5*10^7 5*10^7 is the mass of the rocket times the rocket velocity.And why would I divide 5*10^13 by 2*10^10, when 2*10^10 is what I multiplied the velocity by in the first place? Or is the rocket mass times...

Or it just means we don't have the technology yet to alter the course of something so massive.

So a.) is 5*10^7, b.) is 2.5*10^-3 and c.) is the amount of time to get 10,000 km well if the velocity is 2.5*10^-3 10,000 km = 10,000,000 m 10 million divided by the velocity is the amount of time to move that far. So 10million/2.5*10^-3= 4*10^9 seconds. Am I correct with which answers go...

In my notes for this class, the only related thing I have is a rocket burning fuel. m(dV/dt)= -vex(dm/dt) where vex is the exhaust velocity. The problem i have with this though, is that the mass does change (rocket burning fuel), but you said that was negligible. Then after that step...

Wait, so is 50million the answer to question A? I feel like I'm so close to understanding, but I have no idea how to get the change in velocity in terms of the mass.

So... multiplying the burn speed by the mass of Apophis results in 5*107 What good is this number? edit: Do I set the initial side equal to this value to find the initial velocity of the asteroid?

Those are the two different mass values. What do you mean by "< <"