Recent content by lauriecherie

So I'm getting... .25*PI*g * (The definite integral from 0 to 1.5 of (h + kh^2 + .2 + .2kh). I don't know what to do with this k. If I integrate I'm getting .25*PI*g * { (h^2/2) + ((kh^3)/(3)) + (.2h) + (.1k*h^2)} evaluated from 0 to 1.5

Ok so I can pull out PI, g, and .25? Or just PI and g? And do i distribute (1 + kh) * (h + .2)?

I'm not sure what I need to do. The k is what's causing me problems. I tried factoring but that got me no where. Can you explain what I need to do or explain to me how the book got this answer to this integral? This one of course isn't my problem but can you explain how they intergrated this...

Homework Statement can someone please explain to me how to intergrate this: the definite integral from 0 to PI of (.25*PI) (1+kh) g (h + .2) dh I can leave g, PI, and k in the formula. Homework Equations The Attempt at a Solution

I added the Kinetic energy plus the gravitational potential energy initial. I then set that equal to total mechanical energy final which only included kinetic energy. since I had mass on both sides I canceled out the mass. I was then left with gh + (.5*26^2) = (.5*24^2).

I got 691.16 J for the total mechanical energy initial. For the total mechanical energy final I came out with 288 J. So it was reduced by 288 J?

Homework Statement A 79 kg skier leaves the end of a ski-jump ramp with a velocity of 26 m/s directed 25° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 24 m/s, landing 18 m vertically below the end of the ramp. From the launch to...

Nevermind. It is correct.

Homework Statement A 1.90 g ice flake is released from the ege of a hemispherical bwon whose radius r is 28.0 cm. The flake bowl contact is frictionless. What is the speed of the flake when it reaches the bottom of the bowl. Homework Equations I took KINETIC ENERGY = GRAVITATIONAL...

Solved. Thanks for jogging my memory!

This is something else I've tried. Power = work / time = force (dot product) displacement = force * average velocity * cos(37 desgrees). I cam out with 347. 41 W. Am I doing this correctly?

Is it power?

Homework Statement A 100 kg block is pulled at a constant speed of 3 m/s across a horizontal floor by an applied foce of 145 N directed 37 degrees above the horizontal. What is the rate at which the force does work on the block? _____ W Homework Equations work = force (dot product)...

Homework Statement In the figure below, the coefficient of kinetic friction between the block and inclined plane is 0.17 and angle θ = 55°. What is the block's acceleration (magnitude and direction) assuming the following conditions? (b) It has been given an upward shove and is still sliding...

Ok so that is 21.94 m/s right? I don't know what to do with this info! I am so lost!