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Diano4ka-milaya [45]
3 years ago
11

The relatively long half-life of lipid soluble hormones (steroid hormones) compared to water soluble hormones is due in part to

the way that these hormones are passed into the filtrate from the glomerular capillaries. What's the best explanation for why lipid soluble (steroid) hormones have a relatively long half-life?
Biology
1 answer:
Arturiano [62]3 years ago
4 0

Answer:

The transport protein-hormone complexes of lipid soluble proteins can not cross the filtration membrane of glomerulus.

Explanation:

Steroid hormones are water-insoluble (hydrophobic) in nature. Since blood has watery medium, the plasma, the lipid-soluble hormones travel in blood through transport proteins. The transport protein-hormone complexes are too large to cross the filtration membrane of glomerulus. The impermeability of the filtration membrane of glomerulus for protein-hormone complexes increases the half-life of lipid-soluble proteins.

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Answer:

False

Explanation:

The cells of collenchyma have evenly thickened cell wall

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Answer:

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 7.8 cm from the axis of rotation. (a) Calcul
Novosadov [1.4K]

Given data:

● Angular velocity of the turntable ω = 33 rev/min

Therefore,

ω = 33 rev/min

ω = 33rev/min × 2π rad/rev × 1 min/60 sec = 3.45 rad/s

● The distance of the watermelon seed from the axis of rotation r = 7.8 cm = 0.078 m

● μs = the coefficient of static friction

Section a:

The seed is undergoes circular motion and it is been effected by centripetal acceleration.

ac = rω^2

ac = 0.078 × 3.45^2

ac = 0.9284 m/s^2

Therefore,

the centripetal acceleration of the seed is 9.274 m/s^2

Section b:

If the seed is observed not to slip at the course of the circular motion, then the supplied frictional force given by the seed and surface of turntable would at least be equivalent to the centripetal force working on the seed.

Centripetal Force = Frictional Force

mrω^2 = μsmg

μs = rω^2 /g

μs = 0.078 × 3.45^2

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9.81

μs = 0.09464

Thus,

the coefficient of static friction is 0.09464.

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