Answer:
490.5 N
Explanation:
Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction
F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.
F=kmg=0.5*100*9.81=490.5 N
Answer:
The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.
Explanation:
The engine of the craft provides an upward thrust of 
so that the space craft descends at a constant speed.
This implies that the net force on the space craft is zero.
The upward thrust will be equal to the downward gravitational pull by Callisto.
So the weight of the craft near the vicinity will be 3480 N.
-- From January 15 to February 6 is a period of 22 days.
-- The period of the full cycle of moon phases is 29.53 days.
-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.
-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at <em>Third Quarter</em>. That's the <em>left half of a disk </em>(viewed from the northern hemisphere).
I really doubt because the universe is supposed to be infinite
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
