Recent content by Sniperman724

Or another way to do it is to graph those two points, then find the area under the curve, because the integral of velocity is position(just a fancy word for the area under the curve)

This sound like a coughinelasticcollisioncough problem :)

use the mini-van equation Y=Yo + Vot + 1/2gt^2 and x=vt also, break his initial velocity into components and try to solve it that way

no, both ice dancers start together, then push a part from each other (m1 + m2)v just equals the momentum of the two dancers as a whole because they both initially start out together. so the only two unknown variables are m1 and m2

use the equation (m1+m2)vf=mv1+mv2, this is an inelastic collision problem. Maybe that will help a little more than p=mv

Yea, I would think so, so now that you have the velocity of the ball in the air, you can find the ball's initial acceleration, then plug that into f=ma to find the force that was needed I really hope this helps

If the ball's initial velocity is at an angle of 43.1 degrees and it lands on a flat fairway, then the ball lands at 43.1 degrees as well. I am not sure if that helps at all

First you will need to find the amount of force that is needed to keep that crate in place using the coefficient of friction from the bed of the truck. Then you will need to use that force as a decelerating force which will be used to find the stopping distance. I would think you could use...

you could try to use ft=p where p equals the change in momentum, and f equals the force applied to the ball. You could try to work backwards using kinematics equations to find how fast the ball will be going at the end of its trajectory, then because we neglect air resistance, the ball is...

If the elevator is going down, then the force on the man would be smaller, and vice versa for going up in an elevator.

try splitting up the .2oN force in a direction of 54 degrees south of east into vectors, and use those numbers to solve the rest.

"What does it mean when the collision is "elastic"?" The collision is elastic because both momentum and energy is conserved at the point of the collision. In an inelastic collision, some momentum and evergy is lost. For an inelastic collision, think about a bullet hitting a wooden box, which...

The ball loses a lot more energy while rolling up a frictionless ramp, than when it actually rolls up a ramp. If you use the equation mgh=1/2mv2+1/2I(omega)2 where I equals the moment of inertia of the ball and omega equals (v2/r2) and if you use just mgh=1/2mv2, the ball that is rolling has...

haha, me neither, I've gone on here to find any kind of help that I needed and was able to find it almost every time :)

I agree, do aerospace engineers only work on a small system? or could they also be involved in a much bigger project?