Hello Again! I think the Answer might be 220 m! ( 1/2) ( 21 m/s + 0 m/s) (21 s) = 220 m
Answer:second class lever needs less force to lift a load
Explanation: because the load is between the fulcrum/pivot point and the force. The fulcrum is far away from the force
I will name block a as Ma=5 kg, block b as Mb=10 kg and mass of the pulley M=3 kg and radius as R. Since the system will accelerate in the direction of the block b because it has greater mass, I will take that direction as positive. Both blocks and the pulley have the same acceleration because the slipping on the pulley is neglected. First, the equations of motion:
Mb*g-Tg=MbαR and
Ta-Ma*g=MaαR,
where Ta and Tb are the tensions of the cord, g=9.81 m/s^2 and α is the angular accereration. Also a=αR where a is the acceleration of the system.
Now the equation of rotational dynamics of a solid body:
(Tb-Ta)R=Iα=(1/2)*M*R^2*α, where (1/2)*M*R^2 is the moment of inertia of a disc.
When we input Tb=Mb*g - Mb*α*R and Ta=Ma*g + Ma*α*R from the first two equations into the third we get: (Mb*g - Mb*α*R - Ma*α*R - Ma*g)*R=(1/2)*M*R^2*α.
We solve for α and get: α=(Mb*g-Ma*g)/((1/2)*MR+Mb*R+Ma*R)=2.97 rad/s^2.
We know that a=α*R and we easily get a=0.4455 m/s^2
1 cubic centimeter=1 millimeter
Answer:
19.09 m/s
Explanation:
u = 0, h = 18.6 m
Use third equation of motion
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 9.8 x 18.6
v = 19.09 m/s
