This question is checking to see whether you understand the meaning
of "displacement".
Displacement is a vector:
-- Its magnitude (size) is the distance between the start-point and
the end-point, no matter what route might have been followed along
the way.
-- Its direction is the direction from the start-point to the end-point.
Talking about the Earth's orbit around the sun, we can forget about
the direction of the displacement, and just talk about its magnitude
(size).
If we pretend that the sun is not moving and dragging the whole
solar system along with it, then what do we see the Earth doing
in one year ?
We mark the place where the Earth is at the stroke of midnight
on New Year's Eve. Then we watch it as it swings around through
this gigantic orbit, all the way around the sun, and in a year, it's back
to the same point that we marked !
So what's the magnitude of the displacement in exactly one year ?
It's the distance between the start-point and the end-point. But the
Earth came back to the same place it started from, so there's no
separation at all between the start-point and the end-point.
The Earth covered a huge distance in that year, but the displacement
is zero.
Answer:
In the attached link is the answer.
Explanation:
Lamina occupies x² + y² = 14y. Outside circle is x² + y² = 49
To find the mass of lamina, integrate given density function over the region
m = ∫∫D P(x, y) dA
Subtitute x = r cosФ and y = r sinФ in x² + y² = 14y
and x² + y² = 49
x² + y² = 14y
(r cosФ)² + ( rsinФ)² = 14(rsinФ)
r² = 14r sin Ф
x² +y² = 49
r² = 49
r = 7
Cntre mass (-x. -y)
-x= i/m ∫∫D xp(x,y) dA = 1/m∫∫ (r cosФ) p( r, Ф)r
dr dФ
-y = 1/m∫∫D yp(x, y) dA = 1/m ∫∫D (r sinФ) p(r, Ф) r drФ
where m = ∫∫D p(x, y) dA
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s
By looking at diagrams, My guess is c.
If it helped, it would be appreciated if you can mark as brainliest