Tutor profile: Julian G.

Read the following German text: Studententheaterstück - Am nächsten Mittwoch, um 17 Uhr, führt die sechste Klasse des Helmholtz Gymnasiums, Hamlet von Shakespeare auf. Bitte kommen sie 15 Minuten vor Beginn an und lassen sie die Kamera zuhause, denn die siebte Klasse wird Foto- und Videographie übernehmen. Answer the following questions: (1) What is being announced? (2) When will it take place? (3) When should the audience arrive? (4) Is photography permitted? Why/Why not?

(1) The sixth grade class of the Helmholtz Gymnasium will perform Hamlet by Shakespeare. (2) Next Wednesday at 5pm. (3) 15 minutes early (4:45 pm). (4) Cameras should be left at home because the 7th grade class will be taking pictures and video.

A air tanker with traditional airfoil wings is in level flight and jettisons fire retardant reducing its flight mass by 50%. By how much must the pilot reduce the tanker's velocity in order to remain in level flight at constant altitude without adjusting any other control surfaces?

In order to remain in level flight at constant altitude without the use of other control surfaces, the pilot must reduce the lift generated by 50% since the aircraft's mass is reduced by 50% following the jettison of fire retardants. The lift equation is: $$L = \frac{1}{2} C\rho v^2 A$$ Where C = Constant of complex dependencies $$\rho$$ = air density v = velocity A = wing area since we want $$L' = \frac{1}{2} L$$, we can see that $$v_f = \frac{1}{\sqrt{2}}v_0$$

One block of mass 'M' rests on a frictionless, level table. It is connected by a massless string, over a pulley at the edge of the table, to a hanging block of mass 'm'. Derive the equations of motion for the masses and find the displacement and velocity at time t..

First we draw the free body diagram for each mass. This helps us use Newton's Second Law: (1) $$\sum F = m a$$ (2) $$T = M a$$ and (3) $$m g-T = m a$$ where T is the tension force resulting from the string between the masses. Next we can substitute the expression for T from (2) into (3), giving: (4) $$m g - M a = m a$$ And then solving for a results in: (5) $$a = \frac{m g}{m+M}$$ Now that we know the acceleration of each block, we can use the equation of motion: (6) $$s = s_0 + v_0 t + \frac{1}{2} a t^2$$ where $$s_0 = 0, v_0=0$$, resulting in: (7) $$s = \frac{1}{2} a t^2$$