Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 = 
× 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m
D- if i was right please mark me as a brainliest answer and a thank you
What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m
