Answer:
what's your problem. How may I help you Buddy
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
<span>The moon's orbit around the Earth will advance in one day:
1°
13° correct answer
27°
29°</span><span />
The force is gravitational because when something is falling is call gravitational
N2 = 3*n1
T2 = 2*T1
V1 = V2
(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1
Since P2 is 6P1, it is 6 times greater than original pressure
