Recent content by Perseverence

Homework Statement Princeton p 411 #6 A 7 kg ball is dropped from 20m. if the speed just before it hits the ground in 18 meters per second, what is the work done by air resistance?[/B]Homework Equations KE= (mv^2)*1/2 PE=mgh[/B]The Attempt at a Solution Mgh equals 1400 joules. PE equals...

Thank you. Liner kinetic energy needs to be calculated completely differently than rotational kinetic energy. Lol.

Im sorry. The correct circumference in the problem is 0.5 m. So circumference calculation is correct. But the answer isn't off by a degree of 10 anyway :-(

Homework Statement P62 Sterling. #11 A new design for a car is to have a large disc like flywheel within the car storing kinetic energy. The flywheel has a mass of 370 kg witha radius of 0.5 meters and can rotate up to 200 revolutions a second. Assuming all of this store kinetic energy can be...

Ok. I finally understand!. It all cancels out! Wow, that was painful. Thank you so much for sticking with me and helping me understand.

Work of gravity can only be determined by knowing the force of gravity, and the force of gravity could only be determined by mg cosine Theta.

Okay. I was wrong before cosine of theta is h / d. I still don't understand why this would explain why cosine is not used in the original equation

I feel like this question is getting so spread out and granular that I'm getting lost from what I'm trying to understand in the first place

I'm sorry, but this isn't helping me understand

Cos (theta) = d/h

?Cos(theta)? H and d are definitely not perpendicular because d is along a slope.

But it asks for work done by of gravity which does require force of gravity on the incline which has a cosine theta component.

Homework Statement If the height of a frictionless incline is h. Determine the work done by the force due to gravity F as the crate of the mass m slides down the incline Homework Equations W=Fd cos (theta ) Force due to gravity perpendicular to incline= mg cos(theta) The Attempt at a...

if you are in a helicopter and something in the air, when it reaches as high as it will go, it's Vertical Velocity will be zero. That is the distance from the helicopter Oh my gosh You're right, I was calculating the wrong distance! I figured out what I was doing wrong. Thank you so much for...

Homework Statement ( p 49 sterling) Ignoring air resistance, if it takes 16 seconds for a package to strike the ground, how high above the ground was a package when it was thrown upward from the stationary helicopter at 15 m/s per second[/B]Homework Equations V=Vo+at D=Vot+(1/2)at^2 The Attempt...