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David F.
Junior TA at UST-ZC
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Physics (Newtonian Mechanics)
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Question:

Birds of prey typically rise upward on thermals. The paths these birds take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume that a bird completes a circle of radius $$6.00\, m$$ every $$5.00\, s$$ and rises vertically at a constant rate of $$3.00 m/s$$. Determine $$(a)$$ the bird’s speed relative to the ground; $$(b)$$ the bird’s acceleration (magnitude and direction); and $$(c)$$ the angle between the bird’s velocity vector and the horizontal.

David F.
Answer:

$$(a)$$ Take the $$+y$$ direction to be upward and the $$+x$$ direction to be along the radius of the circle the bird makes while rising up. In this problem, we choose the $$+x$$ axis to be rotating with the bird with the same constant angular velocity. So, the bird has velocity components: $$v_x=\frac{2\pi R}{T}=\dfrac{6\times2\pi}{5}=7.54\,m/s$$ $$v_y=3\,m/s$$ So, the bird's speed relative to the ground is: $$v=\sqrt{v_x^2+v_y^2}=\sqrt{7.54^+3^2}=8.115\,m/s$$ $$(b)$$ The bird is rising up vertically with constant speed so $$a_y=0$$. The bird has only the radial acceleration that results from his rotation in a circle: $$a=a_{rad}=\dfrac{v_x^2}{R}=\dfrac{7.54^2}{6}=9.48\,m/s^2$$ It is always directed to the center of a circle that lies in a horizontal plan at any instant while the bird is rising up $$(c)$$ $$\theta=\tan^{-1}\left(\dfrac{v_y}{v_x}\right)=\tan^{-1}\left(\dfrac{3}{7.54}\right)=21.7^{\circ}$$

Physics (Electricity and Magnetism)
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Question:

A particle with charge $$+4.20\, nC$$ is in a uniform electric field $$E$$ directed to the left. The charge is released from rest and moves to the left; after it has moved $$6.00\,cm$$, its kinetic energy is $$+2.20\times 10^{-6}\, J$$. What are $$(a)$$ the work done by the electric force, $$(b)$$ the potential of the starting point with respect to the end point, and $$(c)$$ the magnitude of $$E$$ ?

David F.
Answer:

$$(a)$$ From the work-energy theorem, Work done by the net force on a particle equals the change in the particle’s kinetic energy: $$W_{a\to b}=\Delta K=K_2-k_1=2.20\times 10^{-6}\, J-0$$ $$\therefore W_{a\to b}=2.20\times 10^{-6}\, J$$ $$(b)$$ $$V_a-V_b=\frac{W_{a\to b}}{q}=\frac{2.20\times 10^{-6}}{4.2\times10^{-9}}=523.81\, V$$ $$(c)$$ Because the electric field in this problem is uniform, we use the following formula: $$E=\frac{V_a-V_b}{s}=\frac{523.81}{6\times10^{2}}=8.73\times10^3\, V/m$$

Calculus
TutorMe
Question:

Evaluate the following integral: $$\int\frac{4x^3}{(x^4+1)^2} dx$$

David F.
Answer:

We can evaluate this integral using the following substitution: Let $$u=x^4+1$$ $$\to$$ $$\dfrac{du}{dx}=4x^3$$ $$\therefore dx=\dfrac{du}{4x^3}$$ We substitute this back into the integral to get $$\int\frac{4x^3}{u^2} \frac{du}{4x^3}=\int\frac{du}{u^2}=-\frac{1}{u}+c$$ Finally, we substitute for $$u=x^4+1$$ $$\int\frac{4x^3}{(x^4+1)^2} dx=-\frac{1}{x^4+1}+c$$

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