The scheme whereby occupants in a pair of shuttles is as follows
use a strong cable with large weight on the end
Then use the orbital naneuvering system(OMS) to set the whole work as spinning about their common center of gravity.
If dirt and grease were good conductors of electrical current, then we could make wire
out of dirt and grease instead of expensive copper. Sadly, they're not. So a coating of
dirt and grease on the wire can have a substantial impact on the connection, and can
even block the flow of current across the connection completely. Moreover, in the case
where the ends of the wires are to be soldered, solder does not adhere to dirty wire.
Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = 
let's calculate
v = 
v = 
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force
by the worker must be equal to the friction force
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2

b. The work is done on the crate by this force is the product of its force
and the distance traveled s = 4.35

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
