Recent content by ice ace

i think easier way is to use kinematics equation : vf^2=vi^2+2ad then add 1.1m with d to get it

use right angle triangle to find hypoteneuse using 17 and 10 as the sides. Then calculate the angle (so between the ground and the rope = between side length 10 and the hypoteneuse) using trig. You can subtract the length of the hypoteneuse by the length of the rope, then use that length with...

yes I think that's a way to do it, maybe typo on calculator. I got 1.186m

I think the problem is how you add the numbers together. Since the test charge placed in middle is positive, the negative charge will try to attract it and and positive charge will repel it. When you add the two values of E together as you did, you are cancelling the effect of the negative...

Maybe I'm wrong, but I know that v= 0.45c in that axis that the Earth and spaceship travel in, but v in the perpendicular axis is 0, is it not?=>spaceship is not moving in the perpendicular axis with respect to earth.

Since the beacon is fired in perpendicular direction, vperpendicular= 0.65c with respect to rocket, and since the rocket is moving at 0m/s in perpendicular axis with respect to earth, applying lorentz transformation in that perpendicular axis will still give beacon velocity of 0.65c

velocities are vector quantities.. so 17-(-30) because the ball is changing direction then divide by the time

Hi, I did something like that last year in high school. At low velocity, frictional force increases linearly with speed, quadratically at higher speeds (eg. 50km/hr). I'm not positive from my memory but the F friction = crosssectional area*constant depending on the shape / roughness of...