The <em>estimated</em> displacement of the center of mass of the olive is
.
<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>
In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> (
) and <em>static</em> conditions (
) to estimate the displacement of the center of mass of the olive (
):
(1)
Where:
- x-Coordinate of the i-th element of the system, in meters.
- y-Coordinate of the i-th element of the system, in meters.
- x-Component of the net force applied on the i-th element, in newtons.
- y-Component of the net force applied on the i-th element, in newtons.
- Mass of the i-th element, in kilograms.
- Gravitational acceleration, in meters per square second.
If we know that
,
,
,
,
,
and
, then the displacement of the center of mass of the olive is:
<h3>Dynamic condition
![\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]](https://tex.z-dn.net/?f=%5Cvec%7Br%7D%20%3D%20%5Cleft%5B%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%2B%280%29%5Ccdot%20%280%29%20%2B%20%281%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%20%2B%20%281%29%5Ccdot%20%28-3%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%20%2B%200%20%2B%20%281.50%29%5Ccdot%20%289.807%29%2B%28-3%29%7D%2C%20%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%20%2B%20%280%29%5Ccdot%20%283%29%20%2B%20%282%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%20%2B%282%29%20%5Ccdot%20%28-2%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%20%2B%20%283%29%2B%281.50%29%5Ccdot%20%289.807%29%2B%28-2%29%7D%20%20%5Cright%5D)
![\vec r = (0,704, 1.233)\,[m]](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%20%280%2C704%2C%201.233%29%5C%2C%5Bm%5D)
</h3>
<h3>Static condition</h3><h3>
![\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]](https://tex.z-dn.net/?f=%5Cvec%7Br%7D_%7Bo%7D%20%3D%20%5Cleft%5B%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%20%2B%20%281%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%20%2B%20%281.50%29%5Ccdot%20%289.807%29%7D%2C%20%5Cfrac%7B%280%29%5Ccdot%20%280.50%29%5Ccdot%20%289.807%29%20%2B%20%282%29%5Ccdot%20%281.50%29%5Ccdot%20%289.807%29%7D%7B%280.50%29%5Ccdot%20%289.807%29%2B%281.50%29%5Ccdot%20%289.807%29%7D%20%20%5Cright%5D)
</h3><h3>
![\vec r_{o} = \left(0.75, 1.50)\,[m]](https://tex.z-dn.net/?f=%5Cvec%20r_%7Bo%7D%20%3D%20%5Cleft%280.75%2C%201.50%29%5C%2C%5Bm%5D)
</h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

![\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]](https://tex.z-dn.net/?f=%5Coverrightarrow%7B%5CDelta%20r%7D%20%3D%20%280.704-0.75%2C%201.233-1.50%29%5C%2C%5Bm%5D)
![\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]](https://tex.z-dn.net/?f=%5Coverrightarrow%7B%5CDelta%20r%7D%20%3D%20%28-0.046%2C%20-0.267%29%5C%2C%5Bm%5D)
The <em>estimated</em> displacement of the center of mass of the olive is
. 
To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931
Answer:
v = 4.4 m / s
Explanation:
Unfortunately, the exercise scheme does not appear. Let's analyze the problem the marble leaves point A with an initial velocity, goes down and then rises to a given height where its velocity is zero, in the whole trajectory they tell us that the resistance is zero, so we can use the conservation relations of the enegy.
Starting point. Point A
Em₀ = K + U = ½ m v2 + mg y_a
point B.
Em_f = U = m g y
the energy is conserved
Em₀ = Em_f
½ m v² + mg y_a = m g y
½ m v² = m g (y -y_a)
v =
In the exercise the diagram is not seen, but the height of point A must be known, suppose that y_a = 4 m
v =
v = 4.4 m / s
There may be an esoteric technical shade or nuance of difference. But I've been an electrical engineer for 40 years now, and have always used them interchangeably.
(I would have answered your question by saying "No.", but this website won't accept an answer that's less than 200 characters long.)
True. All scientific inquiry processes begin with a question.
Hope this helped! Please give me Brainliest!:)