Posted by Lauren on Monday, July 2, 2012 at 12:53pm.
KE gained = PE lost
(1/2)mv^2 = mgh
v = sqrt(2gh)= sqrt(2*9.8*0.54)m/s
First of all convert 54cm to m
1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9
First of all convert 54cm to m
1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9
First of all convert 54cm to m
1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9
First of all convert 54cm to m
1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9
First of all convert 54cm to m
1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9
Related Questions
physics - A novice skier, starting from rest, slides down a frictionless 33.0&...
Physics - A novice skier, starting from rest, slides down a frictionless 35.0 ...
Physics - A novice skier, starting from rest, slides down a frictionless 24.0 ...
physics - A novice skier, starting from rest, slides down a frictionless 31.0&...
Physics - A novice skier, starting from rest, slides down a frictionless 35.0 ...
physics - A 0.50-kg block, starting at rest, slides down a 30.0° incline ...
physics - A 0.50-kg block, starting at rest, slides down a 30.0° incline ...
physics please help :/ - A 0.50-kg block, starting at rest, slides down a 30.0&...
physics - A frictionless30degrees incline should provide an acceleration of 4....
physics - a mass is released from rest from a height of 4m. it slides down a ...
For Further Reading