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Tutor profile: Paul H.

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Paul H.
Former assistant Judo instructor and Boy Scout mentor
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Questions

Subject: French

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Question:

Decrivez votre chambre en utilisant le vocabulaire appris en cours. (Negligez les accents)

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Paul H.
Answer:

Ma chambre est tres petite. Mon lit est a cote de mon etagere. J'ai beaucoup de livres dedans. En face de l'etagere j'ai un bureau. C'est la que je fais mes devoirs. En-dessous de mon lit j'ai des vetements et mon materiel de sport.

Subject: Calculus

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Question:

Two trains leave from the same point. One leaves at noon going east at a constant 10mph. The other leaves at 1 pm going south at a constant 15mph. How fast is the distance between the two trains changing at 5 pm?

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Paul H.
Answer:

Let c be the distance between the trains. At any time t, using the Pythagorean theorem, the distance between the trains can be expressed as: c^2=(10 t)^2+(15 (t-1))^2 (The "t-1" term is due to the fact that the second train leaves an hour later than the first) To obtain the rate of change, we'll take the time derivative of both sides. d/dt [c^2] = d/dt [(10 t)^2+(15 (t-1))^2] using the chain rule, 2c * d/dt[c] = 2*(10 t) * d/dt[10 t] + 2*(15 (t-1)) * d/dt[[15 (t-1)] c * d/dt[c] = (10 t) * 10 + (15 (t-1)) * 15 c * d/dt[c] = 100 t + 225(t-1) d/dt[c] = {100 t + 225(t-1)}/c at t=5, c^2=6100 miles so c=78.10 miles Plugging in these values, d/dt[c] = {100*5 + 225(5-1)}/78.10 d/dt[c] =17.9 mph

Subject: Physics

TutorMe
Question:

There is a solid sphere of radius R and mass m at the top of a slope of angle 60 degrees. The sphere initially at rest, begins to roll down the hill in a straight line without slipping. Ignore the dragging force on the sphere due to friction. A) What is the kinetic energy of the sphere after it rolls a distance d? B) What is the speed of the sphere at this time?

Inactive
Paul H.
Answer:

A) Due to the conservation of mechanical energy, the potential energy of the sphere at the top of the slope will be equal to the kinetic energy of the sphere after it travels a distance d. We, therefore, need to calculate the potential energy of the sphere at the top of the slope. Potential energy U=mgh We can set h to be the difference between the initial height of the sphere at the top of the slope and its height after rolling a distance d. Therefore, h=dsin(60) Because of conservation of energy, U(top)=KE(bottom). KE(bottom)=mghsin(60) B) We can calculate the speed of the sphere from the kinetic energy. However, the sphere's total kinetic energy is equal to the sum of translational kinetic energy and rotational kinetic energy. KE(bottom)= 1/2 m v^2 + 1/2 I ω^2 Since the sphere does not slip, we can easily express ω in terms of v. Simply, v=Rω Thus, KE(bottom)=1/2 v^2 (m+I/(R^2))=mgdsin(60) Thus, v^2=2mgdsin(60)/[m+I/(R^2)] The moment of inertia of a solid sphere is I=2m(R^2)/5 v=Sqrt{2mgdsin(60)/[7m/5]} =Sqrt{10gdsin(60)/7}

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