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HACTEHA [7]
4 years ago
14

Prolonged use of lithium to treat bipolar disorder can cause damage to the kidneys such that epithelial cells of the collecting

duct cannot insert aquaporin channels in their membranes. Which one of the following symptoms would you least expect in a person with such kidney damage?A. polyuria (excessive urine production)B. polydipsia (frequent drinking)C. excessive thirstD. high levels of vasopressin secretionE. glucosuria (glucose in the urine)
Biology
1 answer:
rusak2 [61]4 years ago
7 0

Answer:

E. glucosuria (glucose in the urine)

Explanation:

Generally, glycosuria occurs in patients with kidney changes due to diseases such as Wilson's disease or cystinosis, can also be a hereditary problem, but is not expected in patients with kidney damage caused by prolonged lithium use.

Normally, the kidneys filter the blood, eliminating all substances that are not necessary for the body to function, while glucose is reabsorbed in the blood because of its importance in energy production, but people with renal glycosuria do not reabsorb glucose. , which causes it to be eliminated in the urine, occurring glucosuria.

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A 20-pound object is dropped from a 50-foot bridge onto the ground below it. A 50-pound object is dropped from a 100-foot cliff
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The 50 pound object will reach the ground first.

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Using the second equation of motion, we can derive the time taken for each of the object to reach the ground. As the object is thrown from the top of the bridge or cliff, acceleration due to gravity will be acting on them.

So, s = ut+\frac{1}{2} at^{2}

The value of s for the object thrown from 50 foot bridge is given as 50 ft. The initial velocity of the object will be zero. The acceleration can be written as the ratio of mass to gravitational force.

Then, s = (0) t +\frac{1}{2}*\frac{m}{F}*t^{2}

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Then, consider the time taken for the 50 pound object to reach the bottom of 100 ft cliff as t₁ and the displacement in this case is 100 ft.

Then, 100 = (0)t +\frac{1}{2}*\frac{50}{F}*t_{1}^{2} \\\\t_{1}^{2} =\frac{200F}{50}

t_{1} =  2 \sqrt{F}

Since, the gravitational force acting on both the objects will be same, the ratio of time taken is

\frac{t }{t_{1} }=\frac{\sqrt{5F} }{2\sqrt{F}} = \frac{\sqrt{5} }{2}

So, the time taken for the 20 pound object to reach the ground of 50 ft bridge is √5 s =2.236 s and the time taken for the 50 pound object to reach the ground of 100 ft cliff is 2 s.

Thus, the 50 pound object will reach the ground first.

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