TBR exam 3 questions

[wolverine1]

Question on exam 3, question 52 on PS.

I can't for the life of me figure out how C23 is in series for C1. Any ideas?

Also, for number 41 on PS, I don't see how answer choice D is the correct choice. When I look at the graph, it clearly shows that it is releasing the least heat from 1000K, and then progressively releases more heat.

Finally, number 25 of PS. Why do we have to plug the mass formula into the velocity one? Why can't we plug the mass value directly into the velocity formula?

Thanks,

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[Morsetlis]

You would get a higher yield actually posting the questions and not their references.

 

[BerkReviewTeach]

Question on exam 3, question 52 on PS.

I can't for the life of me figure out how C23 is in series for C1. Any ideas?

Also, for number 41 on PS, I don't see how answer choice D is the correct choice. When I look at the graph, it clearly shows that it is releasing the least heat from 1000K, and then progressively releases more heat.

Finally, number 25 of PS. Why do we have to plug the mass formula into the velocity one? Why can't we plug the mass value directly into the velocity formula?

Thanks,

For #52, to pass through either C2 or C3, you must first pass through C1, so C1 is in series with both C2 and C3.

For #41, if you look at the lines for any of the gases, they either are flat or have a positive slope (on the heat capacity versus temperature graph). That means that as you raise the temperature of the material, its heat capacity is increasing. This tells us that more heat is needed to raise the material from 1000K to 1001 K than from 999K to 1000K. It also tells us that less heat will be released when the material drops from 1000K to 999K than it releases when it drops from 1001K to 1000K. This is what choice D is telling us.

For #25, because mass affects the speed of the particle (given the same KE, a lighter particle has greater speed) and the equation for the radius of curvature shows that r is proportional to both mass and velocity, a change in mass will affect the radius in two ways (1) directly with any chance in mass and (2) indirectly by changing the incident velocity of the particle when it enters the perpendicular magnetic field.

If the magnitude of the velocity didn't depend of the mass of the particle, then you could just directly substitute mass into the equation.