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zimovet [89]
3 years ago
9

A runner produced hypotonic sweat while running a marathon in hot weather. Aner the race, he drank large volumes of water. As a

result of the water intake, his body cells will ________.
A) shrivel.
B) swell.
C) crenate
D) shrink
E) not change.
Biology
2 answers:
alekssr [168]3 years ago
8 0

Answer:

B) swell.

Explanation:

Secretion of sweat is a way the body helps to maintain the homeostasis condition of the body. Sweat comprises of majorly water and electrolytes(e.g Sodium, chloride, calcium, potassium etc).  A sweat is said to be hypotonic when it has a lower concentration of electrolytes than the cells of the sweat glands. These sweat glands primarily secrete sweat in response to nervous stimuli (e.g exercise) or emotional stimuli. So when large amount of water is drank by a  runner who produced hypotonic sweat, his body cells will swell in response to the water intake.

choli [55]3 years ago
4 0

Answer:

Option B) Swell

Explanation:

When runner was running he was sweating which results in loss of water from cells and cells become shrink. But at the end of marathon when he drink water his cells will gain water and store it, which results in swallowing of cells.

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Answer:

Distance Below the top = 6,02 m

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(1) V_{1f} =V_{1O}  - gt

As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

We need the answer when h=0, and from the initial conditions V_{1O} = 0, h_{1O} = 24m, you get:

(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

Replacing this time in (2), the final velocity is:

(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

h_{cross} = 17,98 m

To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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