<h2>Explanation:</h2><h3>3. </h3>
When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:
- Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.
_____
<h3>4.</h3>
The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...
n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33
- none of the offered choices is correct. The closest is 1.34.
Answer:
<h2>Angular Displacement 6.28 radians</h2>
Explanation:
for circular motion we are expected to solve for Angular Displacement it is measured in radian
Measurement of Angular Displacement.
we can measure it using the following relation
∅= s/r
where
s = the distance travelled by the body, and
r = radius of the circle along which it is moving.
given that
circumference c, s= 400 m
r= ?
we have to solve for the radius
we know that circumference

400= 2*3.142*r
400= 6.282*r
divide both sides by 6.284 we have
400/6.284
r= 63.63 m
Angular displcament
∅= 400/63.63
∅= 6.28 radians
I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps
Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F - 
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F -
= m x a
F -
= 50 x 0
F -
= 0
F =
-------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force
is opposing the movement of the box.
∑F = 1.5F - 
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F -
= m x a
1.5F -
= 50 x 0.1
1.5F -
= 5 ---------------------(iii)
<em>Substitute </em>
<em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
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