Answer:
Mescarinic and Nicotinic
Explanation:
Postganglionic fibers can be present in both sympathetic and parasympathetic divisions, their main difference resides in how in the sympathetic division the postganglionic fibers are adrenergic and use norepinephrine (noradrenalin) as a neurotransmitter, in the parasympathetic division, on the other hand, fibers are cholinergic and use acetylcholine as a neurotransmitter, the<em> postganglionic neurons of sweat glands release acetylcholine for the activation of muscarinic receptors, another kind of receptor for acetylcholine are nicotinic receptors </em>that act as transmembrane sodium/potassium channels, while muscarinic receptors need to act through intracellular proteins.
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I will use [pV/T] in the state 1 = [pV/T] in the state 2.
State 1:
p = 1.0 atm
V = 25 liter
T = 100 + 273.15 = 373.15 K
State 2:
p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm
V= ?
T = 25 + 273.15 = 298.15 K
Application of the formula
1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K =>
V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter
Answer:
<h2>Dog's mitochondria lack the transport protein that transport pyruvate ( end product of glycolysis) across the outer mitochondrial membrane
.</h2>
Explanation:
1. As given here that dog's mitochondria can use only fatty acids and also amino acids for their respiration, and as compared to others, Dong's cell produce more lactate then normal, this indicate that his mitochondrial membrane is different then others.
2. The aerobic phases of cellular respiration in eukaryotes occur within mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.
3. These condition shows that dog's mitochondria lack the transport protein of mitochondria that moves pyruvate across the outer mitochondrial membrane.
Explanation:
(a) The given data is as follows.
Pressure on top (
) = 140 bar = 
(as 1 bar = 
)
Temperature = 
= (15 + 273) K = 288 K
Density of gas = 
= 0.4548
= 
= 
Hence, pressure at the natural gas-oil interface is .
(b) At the bottom of the tank,
= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]
= 
= 309.8 bar
Hence, at the bottom of the well at pressure is 309.8 bar.
