Explanation:
Let us assume that forces acting at point B are as follows.

= 0 ...... (1)
= 0
= 0 .......... (2)
Hence, formula for allowable normal stress of cable is as follows.

T = 
= 3925 kip
From equation (1),
= -3925
= -3925
= 12877.29 kip
From equation (2), -12877.29 (Cos 60) + W = 0
= 0
W = 6438.64 kip
Thus, we can conclude that greatest weight of the crate is 6438.64 kip.
Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F =
q =
we calculate
q =
q =
Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer
(C).
When there is an angle between the two directions, the cosine of the angle must be considered.
Step by step Solution
The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.
The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.
The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is
.
Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the
meaning that there is no work done in the direction perpendicular to the motion.
The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is 
Answer:
L = mp*v₀*(ms*D) / (ms + mp)
Explanation:
Given info
ms = mass of the hockey stick
uis = 0 (initial speed of the hockey stick before the collision)
xis = D (initial position of center of mass of the hockey stick before the collision)
mp = mass of the puck
uip = v₀ (initial speed of the puck before the collision)
xip = 0 (initial position of center of mass of the puck before the collision)
If we apply
Ycm = (ms*xis + mp*xip) / (ms + mp)
⇒ Ycm = (ms*D + mp*0) / (ms + mp)
⇒ Ycm = (ms*D) / (ms + mp)
Now, we can apply the equation
L = m*v*R
where m = mp
v = v₀
R = Ycm
then we have
L = mp*v₀*(ms*D) / (ms + mp)