Answer:
Homeostasis refers to the body's ability to maintain a stable internal environment (regulating hormones, body temp., water balance, etc.). Maintaining homeostasis requires that the body continuously monitors its internal conditions.
Explanation:
Answer:
Homeostasis refers to the body's ability to maintain a stable internal environment.
Explanation:
Maintaining homeostasis requires that the body continuously monitors its internal conditions.
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(b) Electrical-->Chemical-->Chemical
(c) Chemical-->Electrical-->Chemical
(d) Electrical-->Chemical-->Electrical
Answer:
(d) Electrical-->Chemical-->Electrical
Explanation:
A nerve impulse is the transmission of an electrical change along the neuron's membrane from the point at which it is stimulated (synapse). The normal direction of impulse in the body is from the cell body to the axon. This nerve impulse, or action potential, is a sudden and rapid change in the transmembrane potential difference.
Normally, the membrane of the neuron is polarized at rest, which means that the ionic constitution of the medium internal to the membrane is different from the external medium, which generates different electrical charges in one medium and the other, so this difference, ie , the potential during rest is negative (-70 mV). The action potential thus consists of a rapid reduction of membrane negativity to 0mV and inversion of this potential to about + 30mV, followed by a rapid return to values slightly more negative than the resting potential of -70mV.
Nervous impulse or action potential, therefore, is a phenomenon of an electrochemical nature and occurs due to changes in the permeability of the neuron membrane. These permeability modifications allow ions to pass across the membrane. Since ions are electrically charged particles, changes also occur in the electric field generated by these charges.
Thus, we can say that the correct answer to this question is: Electrical -> Chemistry -> Electrical
Answer:d
Explanation:
The following life cycle stages do both complete and incomplete metamorphosis of insects have in common is adult. Thus, option B is correct.
Larva" is a stage in the life cycle of an insect among the following choices given in the question that a caterpillar represents. A Caterpillar represents the larva stage in the life cycle of an insect. The Caterpillar is most specifically Butterfly Larva. In this stage of the Butterfly life cycle, the Caterpillar spends most time eating, growing and shedding their exoskeleton.
Life cycle stages do both complete and incomplete metamorphosis of insects have in common is adult. A Caterpillar represents the larva stage in the life cycle of an insect. The Caterpillar is most specifically Butterfly Larva. Larva" is a stage in the life cycle of an insect among the following choices given in the question that a caterpillar represents.
Therefore, The following life cycle stages do both complete and incomplete metamorphosis of insects have in common is adult. Thus, option B is correct.
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Surface area to volume ratio, in simple means the size of surface area to the volume of substance that can pass through it at a particular time.
Amoeba and some bacteria are flat and have large surface area to volume ratio. So the diffusion rate is very high due to large surface area.
Where as humans have small surface area: volume so diffusion is very slow or does not take place at all.
b. 74.800
с. 1.55
Answer:
The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.
Explanation:
From the given information:
For food source A; we have:
3P₁ + P₂ + 2P₃ = 58 units of food A ---- (1)
For food source B; we have:
2P₁ + 4P₂ + 2P₃ = 70 units of food B ---- (2)
For food source C; we have:
P₁ + P₂ = 20 units of food C ----- (3)
From equation (1) and (2); we have:
3P₁ + P₂ + 2P₃ = 58
2P₁ + 4P₂ + 2P₃ = 70
By elimination method
3P₁ + P₂ + 2P₃ = 58
-
2P₁ + 4P₂ + 2P₃ = 70
P₁ - 3P₂ + 0 = - 12
P₁ = -12 + 3P₂ ---- (4)
Replace, the value of P₁ in (4) into equation (3)
P₁ + P₂ = 20
-12 + 3P₂ + P₂ = 20
4P₂ = 20 + 12
4P₂ = 32
P₂ = 32/4
P₂ = 8
From equation (3) again;
P₁ + P₂ = 20
P₁ + 8 = 20
P₁ = 20 - 8
P₁ = 12
To find P₃; replace the value of P₁ and P₂ into (1)
3P₁ + P₂ + 2P₃ = 58
3(12) + 8 + 2P₃ = 58
36 + 8 + 2P₃ = 58
2P₃ = 58 - 36 -8
2P₃ = 14
P₃ = 14/2
P₃ = 7
Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.
Answer:
The aerobic phase produces a net of 28-30 ATP.
Explanation:
because you are using more oxygen.
OA 3.125 x 108 atoms of calcium-41 and 4.375 x 10⁹ atoms of potassium-41
OB. 6.25 x 108 atoms of calcium-41 and 4.6875 x 10° atoms of potassium-41
OC. 6.25 x 108 atoms of calcium-41 and 4.375 x 109 atoms of potassium-41
OD. 3.125 x 108 atoms of calcium-41 and 4.6875 x 10° atoms of potassium-41
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The correct answer for this question will be option D.
Calcium-41 = 3.12× atoms
Potassium-41 = 4.69× atoms
As stated in the question,
Half life of isotope of Ca-14 which decays into K-14 = 1.03 × years
Therefore, after 1.03 × years (1 half life)
Ca-41 will be 50%
and, K-41 will be 50%
After 2.06 × years (2 half lives)
Ca-41 will be 25%
K-41 will be 75%
After 3.09 × years (3 half lives)
Ca-41 will be 12.5%
K-41 will be 87.5%
After 4.12 × years (4 half lives)
Ca-41 will be 6.25%
K-41 will be 93.75%
After 4.12 × years,
Ca-41 = 6.25/100 × 5×
= 3.12 × atoms
K-41 = 93.75/100 × 5×
= 4.69 × atoms
Therefore, option D is correct.
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