FL 9 - 1, 21, 41

I had a couple questions from the physical sciences section of Full-Length 9. The first one is actually #1. I chose answer c, and I still can't understand why the answer is b. Wouldn't the elastic part be easier to stretch? My next question is number 21, I just can't put together an equation-based relation that includes velocity. Also, could you explain number 41? Thank you!!
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Question 1:

For this question, we need to analyze the graph.  The graph is stretch versus time (NOT force), for a system with a viscous component and an elastic component in parallel.  Well, if we give a certain CONSTANT force to a spring over time, what happens?  It stretches until it reaches a certain deformation from equilibrium, and then just stays there.  Think about hanging a mass from a spring – that’s a constant force on the spring which causes deformation until the mass is at translational equilibrium.

What would happen to a viscous object without an elastic force?  Well, it would stretch and stretch and stretch… and keep stretching until (probably) it eventually broke.  But either way, it doesn’t reach equilibrium because it’s not elastic.  So it just keeps stretching.

Why is it asymptotic, then?  Well the spring will only stretch until it’s in equilibrium with the force.  At that point, it won’t stretch anymore.  Thus, the stretch of the viscous component (which should be infinite) is limited by the stretch of the spring.

Question 21:

We’re supposed to relate the force created by the spring with this scenario of a pilot landing a jet.  Well, anytime you see force, think acceleration!  F_net = ma, but F also represents the force of the spring here, and thus F_spring = kx.  So, we can say F_spring = ma.  We’re given one more key piece of information:  the final velocity is zero.  Acceleration is simply change in velocity per time or Δv/t.  Since v_f = 0, we can say that a = (v_f – v_i)/t = (0 - v_i)/t = -v_i/t.  Plugging into our equation, we get F_spring = m(-v_i/t) = -mv_i/t.  Thus, the force is directly proportional to the starting velocity.

Question 41:

The first thing we see when looking at the answers is the fact that it says that one has higher or lower energy than the other, and differences in principle quantum number (5sto 5d, etc.).  So, let’s start with energy.  E = hf, so the higher the frequency, the higher the energy.  Red light has the lowest frequency while violet has the highest, so that means red light has lower energy than violet.  Here, we’re dealing with red (Sr) and green (Ba).  So that means that strontium has a lower energy jump than barium does.

Well, for principle quantum number, where is strontium?  It’s in the 5s grouping.  Barium, however, is 6s.  Where are these electrons jumping to?  Well, it should jump to the next-highest-energy orbital.  If we start at an s orbital, we’ll just jump to p or d.  So we’ll want to see that electrons jumping from 5s to 5p or d have less energy than those jumping from 6s to 6p or d.  And that matches with answer choice (B).