A box slides down a ramp inclined at 41 degrees horizonontal with an acceleration of 1.4 m/s^2. The acceleration of gravity is 9.8 m/s^2. Determine the coefficient of kinetic friction between the box and the ramp.
no ideas need any advice
elevator tension problem
An elevator starts from rest with a constant upward acceleration and moves 1 m in teh first 1.5s. A passenger in the elevator is holding a 9.8 kg is holding a 9.8 kg bundle at the end of a vertical cord. G=9.8 m/s^2. What is the tension in the cord as teh elevator...
at the moment the train's speed is 54 km/h. what is the magnitude of the total acceleration? units m/s^2 r=173 and it takes 18.9 s to slow down from 82 km/h to 37 km/h, the a tangential is .6614 so is it (15)^2/173 + .6614 = total?
are you saying then that because its the same size 6 x 6 that the velocity in each direction is teh same? but also to calculate the velocity in y isn't it just 6/9.8?
no i messed up my answer for 9 was 24.8055 , cause it was in km not m and then divide by the vxi which was .0999649 and got the answer i should have gotten
the equation i used was rf=ri+vit+1/2gt^2 and solved for t then subtracted t from 2.6 being the total time of the parobola, but how can i find the length x between the man and the hoop, now that i know the time it takes to get there?
if the difference in height is .302 and you are given vi and theta can you calculate how long it takes for the ball to reach that height? then subtract that value from the total time the ball would have been in the air for? i know it take 1.3 s to reach max h
correct answer was 248.142 how is that s, cause i had that answer but it was .248142, why did they turn the km in m, shouldn't it not have made a differnece?
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a horizontal shelf of 6 m, then another drop of 4m to the bottom of the canyon, which has a horizontal floor. You kick, a .72 kg rock,giving it an inital horizontal...