See what we can offer to your school
"SimpleStudy just makes sense...”
Get the best plan for your school
10 questions from this quiz
Only tension along its length
Horizontal component FTcosθF_T \cos\thetaFTcosθ
v=2πrTv = \frac{2\pi r}{T}v=T2πr
Tension is greater than mgmgmg
Both magnitude and direction vary
Friction force on the tyres
Increases with square of velocity
Four times as much
Horizontal component of normal force
Speed and radius only
Select your subjects, and get access to A+ resources today.