I think Im gonna have to go with C 6.00 T/s but Im not sure
It is through biopsychological feedback.
A class of chemical called a neurotransmitter is important in the transmission of nerve impulses. Neurotransmitters are packaged by the cell into small, membrane-bound sacs called vesicles. Upon receiving a chemical signal, the vesicles move toward the cell membrane and fuse with it, releasing the enclosed neurotransmitters from the terminal end of the nerve cell.
Answer:
If the final question is; at what velocity will the first block start to move outward in m/s?

Explanation:
The motion have the velocity that will make the block move using:




μ
Resolving:





Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
Answer:

Explanation:
The magnitude of the electrostatic force between two charged objects is

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.
In this problem, we have:
is the distance between the charges
since the charges are identical
is the force between the charges
Re-arranging the equation and solving for q, we find the charge on each drop:
