Answer and Explanation:
The clothing after spinning in the dryer cling together. This is because in the dryer they are rubbed against each other and due to this rubbing, electrons are transferred from one to the other clothes and acquire charge as a result of charging by friction thus producing static electricity.
As the material of the clothes in the dryer is different, clinging will be more.
The sticking of these clothes together is known as Static cling.
In case, the clothing are of same material, the static electricity produced as a result of frictional charging would be less and hence less static cling would occur.
Carbon atoms and hydrogen I hope this helps
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
B I believe but hey we also have Google to double check
Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :



F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.