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Tutor profile: Raíla A.

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Raíla A.
Ph.D. in Physics with Teaching Experience
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

A square trapdoor (side A and mass M), is raised vertically in equilibrium on the hinges when it falls by slight trepidation. Despising friction, what angular velocity will it have gained by hitting the ground?

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Raíla A.
Answer:

U: potential energy K: kinetic energy A: length of one side of the square trapdoor M: the mass of the square trapdoor $\omega$: angular velocity I: the moment of inertia g: gravitational acceleration In this problem, the principle of energy conservation is used: $U = K$. $mgA = \frac{1}{2} I \omega^2$. In this case, for a square trapdoor with the axis of rotation located on one of the hinges, the moment of inertia is $I = \frac{2}{3} m A^2$. Therefore, $g A = \frac{1}{3} A^2 \omega^2$. Answer: $\omega = \sqrt{\frac{3g}{A}}$.

Subject: Advanced Physics (Special Relativity)

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Question:

Two atomic clocks are carefully synchronized. One of them stays in London and the other one is on a plane with an average velocity equal to 250 m/s and later back to London. When the airplane returns, the total time interval measured by the clock on the ground is equal to 4 h. What is the difference between measured time intervals by both clocks and which one indicates a short interval?

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Raíla A.
Answer:

The interval measured by the reference frame at rest on earth (London) is given, according to Lorentz transformations by: $\delta t = \frac{\delta t_0}{\sqrt{1- \frac{u^2}{c^2}}}$, where $\delta t_0$ is the time interval measured in at rest with respect to the airplane. We know that airplane speed is much slower than the speed of light (vacuum), so we will have: $(1-\frac{u^2}{c^2})^{-1/2} \approx 1+ \frac{1}{2}(\frac{u^2}{c^2})$. Using the previous results, we will have: $\delta t \approx \delta t_0 (1+ \frac{1}{2} \frac{u^2}{c^2})$. Therefore, $\delta t_0 = \frac{\delta t}{1+ \frac{1}{2} \frac{u^2}{c^2}}$, $\delta t_0 = \frac{4}{1+ \frac{1}{2}\frac{(250)^2}{(3 \times 10^8)^2}$. Now making $\delta t - \delta t_0$, we have $\delta t - \delta t_0 = \approx \frac{\delta t_0}{2} \frac{u^2}{c^2}$. Therefore, $\delta t - \delta t_0 = 2 (\frac{\frac{(250)^2}{(3 \times 10^8)^2}}{1 + \frac{1}{2}\frac{(250)^2}{(3 \times 10^8)^2}})$, $\delta t - \delta t_0 \\aprox 2 \times 6.94 \times 10^-13 = 1.39 \times 10^{-12} h$. Answer: About 5ns. The clock on the plane indicates a shorter time interval.

Subject: Physics

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Question:

In the race between the hare and the tortoise, the speed of the hare is 30 km/h and the tortoise velocity is 1.5 m/min. The total distance is 600m. The hare runs for 0.5min before stopping for a nap. What is the maximum length of the nap so that the hare doesn't miss the race?

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Raíla A.
Answer:

$\delta x = 600 m$, $v_hare = 30 km/h$, $v_tortoise = 1.5 m/min$ and $\delta t =0.5 min$. We want to know $t_nap$. Since motion is uniform, we use only the equation that gives us the distance traveled as a function of time: $\delta x = v.t$ The total time each animal spends to travel the distance ∆x is given by: $t_hare = \frac{\delta x}{v_hare}$ and $t_tortoise = \frac{\delta x}{v_tortoise}$. Then $t_hare = 72$ s and $t_tortoise = 2.4 \times 10^4$ s. In order for the hare not to lose the race, its travel time plus the nap time must be equal to the tortoise's travel time. Therefore: $t_hare + t_nap = t_tortoise$ $t_nap = 2.393 \times 10^4$ s or $t_nap = 6.647$ h. Answer: The hare can take a 6h 38min 49s nap without losing the race to the tortoise.

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