I'm happy to know that the diagram shows how it's all set up.
If I could see the diagram, then I could probably do a much
better job with an answer. As it is ... 'flying blind' as it were ...
I'm going to wing it and hope it's somewhat helpful.
If the pulley is movable, then I'm picturing one end of the rope
tied to a hook in the ceiling, then the rope passing down through
the pulley, then back up, and you lifting the free end of the rope.
A very useful rule about movable and combination pulleys is:
the force needed to lift the load is
(the weight of the load)
divided by
(the number of strands of rope supporting the load) .
With the setup as I described it, there are 2 strands of rope
supporting the load ... one on each side of the pulley. So the
force needed to lift the load is
(250 N) / 2 = 125 N .
Answer: B) Velocity
Explanation:
Velocity is a vector quantity. Velocities have both magnitude and direction.
Answer: If a star's radial velocity is -50 km/s, the frequency of its light would appear to be higher than its true frequency. We usually say that the star's wavelength is blue shifted.
Explanation:
Radial velocity is defined as the velocity of how an object is seen by the axis of a circle.
Then, if the radial velocity is -50km/s, this means that the radius is decreasing, and then that the star is coming towards the viewer.
as the star is coming towards the viewer, we will see a shorter wavelength, which implies that the frequency would appear to be bigger as it really is.
and this is called a blue shift, because the blue light is the visible light with the biggest frequency, while the red light is the visible light with the smallest frequency, then we have:
If a star's radial velocity is -50 km/s, the frequency of its light would appear to be higher than its true frequency. We usually say that the star's wavelength is blue shifted.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = 
= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
The answer to this question is D.) straight line motion
