Answer:
291.598 N-m
291.6 N-m
Explanation:
Let's first take a look at the free bodily diagrammatic representation.
The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).
Let's first determine our angle θ from the diagram
To find angle θ ; we have :
tan θ = 
tan θ = 
tan θ = 1.333
θ = tan⁻¹ (1.333)
θ = 53.13°
Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.
We have:

where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:




Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m
b) From the second diagram, taking the moment at point B
,
we have:



where Force(F) = 1350 N and θ = 53.13° ; we have:



Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m
Answer:

Explanation:
When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.
According to the Snell's law:

a)

b)


c)

d)

e)

f)


Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
with the same generator, so the only factor for producing
the slectric field is only the speed. The faster the rotational speed of the
generator the greater it produce electric field. So the sequence is 3000 rpm
< 3200 rpm < 3400 rpm < 3600 rpm