Answer:
The following options are true;
a) Electron transfer in the ETC is coupled to proton transfer form the matrix to the intermembrane space
c)The outer membrane of mitochondria is readily permeable to small molecules and hydrogen ions.
d) Electron carriers in the mitochondrial matrix include ubiquinone (coenzyme Q), FMN, and cytochrome c.
f)Electron carriers are organized into four complexes of proteins and prosthetic groups
Explanation:
ETC involves series of steps in which electrons are passed rapidly from one component to the next to the endpoint of the chain, where the electrons reduce molecular oxygen, producing water. This requirement for oxygen in the final stages of the chain can be seen in the overall equation for cellular respiration, which requires both glucose and oxygen.
A complex is a structure consisting of a central atom, molecule, or protein weakly connected to surrounding atoms, molecules, or proteins. The electron transport chain is an aggregation of four of these complexes (labeled I through IV), together with associated mobile electron carriers. The electron transport chain is present in multiple copies in the inner mitochondrial membrane of eukaryotes and the plasma membrane of prokaryotes.
To solve this problem we will use the concepts related to energy conservation. Both potential energy, such as rotational and linear kinetic energy, must be conserved, and the gain in kinetic energy must be proportional to the loss in potential energy and vice versa. This is mathematically
Where,
m = mass
v = Tangential Velocity
= Angular velocity
I = Moment of Inertia
g = Gravity
Replacing the value of Inertia in a Disk and rearranging to find h, we have
Replacing,
Therefore the height of the inclined plane is 5.6m
The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right
<h3>How to determine the net force</h3>
Case 1 (Net force between up and downward force)
- Force up (Fu) = 50 N
- Force down (Fd) = 30 N
- Net force 1 (F1) = ?
F1 = Fu - Fd
F1 = 50 - 30
F1 = 20 N up
Case 2 (Net force between right and left)
- Force right (Fr) = 60 N
- Force left (Fl) = 20 N
- Net force 2 (F2) = ?
F2 = Fr - Fl
F2 = 60 - 20
F2 = 40 N right
SUMMARY
- Net force between up and down = 20 N up
- Net force between right and left = 40 N right
From the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)
Thus, the net force of the system will be 40 N horizontal to the right
Learn more about force:
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