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Tutor profile: Gabriel P.

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Gabriel P.
Multi-disciplinary Tutor
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Questions

Subject: Biochemistry

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Question:

The ATP synthase enzyme features a rotor shaft that extends between the rotor in the membrane and the mushroom shaped catalytic head group. The oustide of ATP synthase also has a second long protein fiber that extends between the stator in the membrane and the head group. Why is this second long protein fiber present?

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Gabriel P.
Answer:

This second long protein fiber attaches the catalytic head group to the rest of ATP synthase, and makes sure that the head group does not rotate.

Subject: Basic Chemistry

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Question:

You are splitting water into hydrogen and oxygen. Suppose the amount of hydrogen gas measured is 3.0 moles. How much oxygen gas is present?

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Gabriel P.
Answer:

The molecular formula of water is H2O, telling us that there is twice as much hydrogen present as there is oxygen. Therefore, we divide the moles of hydrogen by 2 to get 1.5 moles of oxygen gas.

Subject: Calculus

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Question:

Standing on level ground, Paul throws a ball into the air. Assuming the ball is modeled by the function (in feet) s(t) = -32t^2 + 16t (t is in seconds), what will be the maximum height the ball reaches?

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Gabriel P.
Answer:

The ball will reach a maximum height when the derivative of the position function (i.e. the velocity) equals 0. So first, calculate the derivative, set it equal to 0, and solve for t. 1. s'(t) = -64t + 16 0 = -64t + 16 -64t = -16 t = 0.25 s We need to make sure this is a maximum by seeing if the function is concave down (i.e the second derivative, the accerleration, will be negative). 2. Take the second derivative, and see if it's negative. s''(t) = -64 The function is always concave down (negative), so the value at t = 0 for the first derivative (velocity) must be a maxmimum. So far, we have only calcualted the time at which the height is a maximum. We take the caluclated value of t = 0.25s, and plug it into the original position function, knowing that the calculated value will be the maximum height. 3. s(t) = -32t^2 + 16t s(0.25) = -32 * (0.25)^2 + 16 * 0.25 s(0.25) = -2 + 4 = 2 Therefore, the ball reaches a maximum height of two feet.

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