Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
Answer:
The tension in the rope is 49.66 N.
Explanation:
Given that,
Mass of bucket m= 23.0 kg
Diameter = 0.300 m
Mass of rope M= 13.0 kg
Height = 10.5 m
Suppose we need to find the tension in the rope while the bucket is falling.
We need to calculate the acceleration
Using balance equation
..(I)
We need to calculate the tension in the rope
Using formula of tension
....(II)
Put the value of T in the equation (I)
Put the value into the formula
Now, put the value of a in equation (II)
Hence, The tension in the rope is 49.66 N.
106.68 centimetres are in 3.50 feet