Recent content by 777lov3r

oooohh, so we have to find his minimal acceleration , and the tension on the rope in order for it not to break , so it would be a=(fg-ft)/m?but isn't tension unkown?

but it's saying that the timer starts when the rock falls off the cliff of an unknown distance then sound travels back to u..in 3.4s, so would u use a formula 3.4=t(rock falls)+t(sound)?:)

so I attempted this by saying 3.4s=t(rock falls)+t(sound back), then solved and got 2 answers, which were 112.2 m and -11908 m, and I tried seeing if the 112.2 m made any sense when I plugged It in, but it would take the rock more than 3.4 s to reach the bottom, so 112.2 m is impossible...any...

but I'm confused on whether or not to use that mans mass, 75 kg, if I draw a free body diagram then fnet=fg-fn, but fnet =ma, so I isolate for a? my teacher doesn't really help me out very much..

A 75 kg petty third wants to escape from a third story jail window. Unfortunately, a makeshift rope of sheets tied together can support a mass of only 50 kg. What is the minimal acceleration with which the thief must lower himself so he may use the "rope" without breaking it? the variables i...

so I am kind of stumped on this one here... A rock is dropped from a sea cliff and the sound of it striking the ocrean is heard 3.4 seconds later. If the speed of sound is 340m/s, how high is the cliff? - now i got some variables, v1=0 m/s time(total)=3.4s...