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# Tutor profile: Brandon S.

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Brandon S.
PhD Structural Engineering Student
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A 1 kg ball is thrown straight up into the air with a velocity of 25 m/sec. There is a constant air resistance force of 0.5 Newtons acting against the ball the entire time. What is the maximum height the ball reaches relative to where the ball was released?

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Brandon S.

First, we need to find the force acting on the ball during its upward flight. This is done by using Netwon's 2nd Law. Fnet = Fgravity + Fair --> Fnet = m*g + 0.5N --> Fnet = 1kg*(-9.81 m/s^2) + 0.5N --> Fnet = -9.31N We can use the net force acting on the ball to find the net acceleration of the ball using F = m*a. F = m*a --> a = F/m --> a = -9.31 m/s^2 Now we can use the conservation of energy to relate the initial and final potential and kinetic energy of the ball. We need this relationship because the potential energy term includes final height which is what we are solving for. 1/2*m*(v_initial)^2 + m*g*h_initial = 1/2*m*(v_final)^2 + m*g*h_final We know that the initial height is 0 meters and the final velocity is 0 m/s (when the ball is at its peak height). Therefore we can simplify our expression. 1/2*m*(v_initial)^2 + m*g*0 = 1/2*m*0^2 + m*g*h_final --> 1/2*m*(v_initial)^2 = m*g*h_final --> h_final = 1/2*(v_initial^2)/g We know the initial velocity (25 m/s) and we calculated g to be 9.31 m/s^2. So, h_final = 1/2*(25 m/s)^2/(9.31 m/s^2) --> h_final = 33.57 meters Therefore, the maximum height reached relative to where the ball was thrown is 33.57 meters. Check: For the final expression used, let's do a unit analysis to make sure the units are what they are supposed to be. We can ignore all the scalar values here. [m] = [m/s]^2/[m/s^2] --> [m] = [m], we can see that the units match.

### Subject:Civil Engineering

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Question:

In general, which is more likely to undergo larger roof accelerations during an earthquake, a 4 story building with stiff concrete shear walls or a long, 8 story, slender tower built with steel moment frames?

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Brandon S.

The 4 story building with stiff concrete shear walls should undergo larger roof accelerations. Such a building would have a much larger stiffness than a slender tower built with moment frames due to the building being shorter and the use of concrete shear walls instead of steel moment frames. Although not given, the masses of the two structures would be more comparable than the stiffnesses of the structures. This is because concrete shear walls have much more mass than steel moment frames, and it was also given that the 8 story building is slender whereas, in general, 4 story buildings with shear walls tend to not be slender. So even though there are twice as many stories, the shorter building has a much higher mass per floor than the slender building. Since the natural period of a structure is equal to square root of the mass over the stiffness, we can see that the natural period of the shorter building is going to be smaller than that of the slender building. Buildings with shorter periods undergo larger accelerations and smaller deflections than buildings with longer periods. Therefore, the 4 story building will have larger roof accelerations during an earthquake.

### Subject:Algebra

TutorMe
Question:

Divide \$100 among four people so that the second will have twice as much as the first, the third will have \$5 less than the second, and the fourth will receive \$20.

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Brandon S.

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