Recent content by Joe91090

i submitted .44 as an answer and it says I am incorrect Here is the original problem : A cannon with a mass of 2220kg fires a 19.5 kg ball horizontally. the cannonball has a speed of 102m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to rool...

does Va2= 0 ?? if it does then i still get 102.44

so i got 889.14 and 101439 for the kinetic energy of the cannon and cannon ball for part A and i add these together to get 102328 for the kinetic energy of the cannon ball in part B. so i did 102328=1/2(19.5)v^2 i got 102.44 for v is this correct ?

so would i use 1/2mv^2=1/2mv^2 to solve ? never mind because that would still make v2= 0 for the cannonball

i could rewrite it (2220)(0)=(19.5)(v2) Im not 100% sure this is the right equation but part B it says the total energy stays the same, to disregard friction and all other parameter retained the same as in part A

yes i wrote x = velocity and 19.5 kg and 2220 kg are the weights given in the problem

wouldnt the cannon barrel be 0 also since its a part of the rigidly mounted cannon ?

of course ! thanks a lot

Homework Statement An 89 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. the astronaut turns away to look at the Earth and several seconds later is 43.8m behind the ship, at rest to relative to the spaceship. The only way to...

Homework Statement So I know that a cannon weighs 2220 kg and a cannon ball weighs 19.5 kg and that when you fire the cannon at 102 m/s the speed of the cannon is .895m/s in the opposite direction. part B of the problem asks how much faster would the ball travel is the cannon was mounted...

No i miss last class and my professor assigns homework online and it is due tomorrow

the question seems so simple I am just looking for what formula or formulas i would use

ok so I am trying to solve for the force on the cable first. I really have no idea what formula to use all i know is that it is at 32 degrees from the horizontal with a 18 kg mass hanging from the end of it.

I thought F=mg x cos(θ) but now I am thinking that's wrong

would 15.56N be too small for the force on the cable. I did 18cos(32).