Recent content by alex740rider

Solve it, thanks to ehild! You had to use all the numbers in part a to solve it correctly (instead of just 1.02) Then plug it back into a = (2kxsin(angle) - 100) / (100 / 9.8) which you will get 7.7235 m/s^2 Thank you to everyone that tried to help! Again big shout out to ehild!

WOW i guess it did made a difference

2*1060*sin(31) = 1091.880719 100/1091.880719 = .0915850956 1.11-.0915850956= 1.018414904

I have updated all the numbers and kept all of the decimals. Can someone tell me what I'm doing wrong or if there's another equation that I can use?

also what equation did you use for it to be 4% higher?

I have the 2 tension forces upwards and gravity downwards

The numbers for the x and y components or the numbers for x in the equation?

Hi all, My name is Alex and I am currently working on my kinesiology degree. I look forward to learning more about the fundamentals of physics and math and I hope we can all get along here. Thank you, Alex

I've been stuck on part b for several hours now (feels like the whole day). 1. Homework Statement : 2. Homework Equations : 2 k * x sin(θ') = 100/g * a sum(Fy) = F net Ma = 2kxsin(angle) - 100 a = (2kxsin(angle) - 100) / (100 / 9.8) 2*kxsin(θ')-100 / 10.2 3. The Attempt at a Solution : What...