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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.

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.

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.

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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