Three ice skaters meet at the center of a rink and each stands at rest facing the center, within arm's reach of the other two. on a signal, each skater pushes himself away from the other two across the frictionless ice. after the push, skater a with mass ma = 70.0 kg moves in the negative y-direction at 2.00 m/s and skater b with mass m = 85.0 kg moves in the negative x-direction at 2.50 m/s. find the x- and y-components of the 80.0 kg skater c's velocity (in m/s) after the push. hintvcx= m/svcy=/s

To solve the problem it is necessary to apply the conservation equations for the moment, specifically for collision. In addition to that, the concepts of vector velocity are necessary, in which the components are obtained and if the total magnitude is necessary.

Conservation of Momentum equation is given by,

We need to find the speed 3, therefore by readjusting the equation we have,

Therefore the components of the velocity would be,

temp rises, and expansion

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.