Enable contrast version

# Tutor profile: Pedro A.

Inactive
Pedro A.
Masters Student in Electronics Engineering
Tutor Satisfaction Guarantee

## Questions

### Subject:Electrical Engineering

TutorMe
Question:

Why are transformer cores constructed in a laminated fashion?

Inactive
Pedro A.

Transformer core's are not built to have any electrical currents passing through them, however since they form conducting loops subjected to changing magnetic fields, Maxwell's laws (in particular Ampère-Maxwell's law) dictate that currents, called "eddy currents" must be present. Since these currents will heat up the core, and thus hinder the efficient transfer of energy from the primary to the secondary side of the transformer, the designers of such apparatus try to keep them to a minimum. The lamination process inherently increases the resistance of the core, thus reducing the magnitude of the induced currents and the heat produced.

### Subject:Calculus

TutorMe
Question:

Please calculate the following indefinite integral $$\int\exp{(x)}\sin{(x)} dx$$

Inactive
Pedro A.

Well, as complex as this integral appears to be it comes down to applying integration by parts. The basic formula in integration by parts is usually written $$\int u dv = u v - \int v du$$ In this particular problem, it comes down to applying this rule twice as you shall see. Start with the initial integral, and apply integration by parts: $$\int\exp{(x)}\sin{(x)} dx = -\exp{(x)}\cos{(x)} - \int\exp{(x)}(-\cos{(x))}dx$$ It seems like we're back to square one, which is why this question is typically put on exams. $$\int\exp{(x)}\sin{(x)} dx = -\exp{(x)}\cos{(x)} + \exp{(x)}\sin{(x)} - \int\exp{(x)}\sin{(x)} dx$$ The "trick" is realizing that integrals can be treated like simple algebraic entities and, as such, we can isolate the integral we are after on the left-hand side of the equation, sic: $$2 \int\exp{(x)}\sin{(x)} dx = -\exp{(x)}\cos{(x)} + \exp{(x)}\sin{(x)}$$ And we are almost done after dividing both sides of the equation by $$2$$: $$\int\exp{(x)}\sin{(x)} dx = -\frac{1}{2}\exp{(x)}\cos{(x)} + \frac{1}{2}\exp{(x)}\sin{(x)}$$ To conclude, always remember that indefinite integrals define whole families of functions up to a constant factor, so the final answer is $$\int\exp{(x)}\sin{(x)} dx = -\frac{1}{2}\exp{(x)}\cos{(x)} + \frac{1}{2}\exp{(x)}\sin{(x)} + C$$ where $$C$$ is an arbitrary constant.

### Subject:Physics

TutorMe
Question:

Assume a clock's operation is regulated by a pendulum with a length $$L$$, whose oscillation frequency is $$f_1$$. If the clock is raised from a height $$h_1$$ to a height $$h_2$$, and assuming all other operating conditions remain the same, how much will the frequency of the clock vary?

Inactive
Pedro A.

If it is assumed that the angle of oscillation of the aforementioned pendulum is small, usually taken to be under 10 degrees. Then it is a well known fact that the oscillation period of such a pendulum is approximately equal to $$T=2 \pi\sqrt{\frac{L}{g}}$$. From this, and given the fact $$L$$ doesn’t change, it may be inferred that the change in the period of oscillation can only come from a change in the gravitational acceleration itself. In lieu of this, if we let $$g_1$$ be the acceleration the height $$h_1$$ and $$g_2$$ be the acceleration at height $$h_2$$, it immediately follows that: $$T_1 = 2\pi\sqrt{\frac{L}{g_1}}$$ and, by the same reasoning: $$T_2 = 2\pi\sqrt{\frac{L}{g_2}}$$ Let $$\frac{T_1}{T_2}=k$$, this implies that: $$\frac{T_1}{T_2} = \frac{2\pi\sqrt{\frac{L}{g_1}}}{ 2\pi\sqrt{\frac{L}{g_2}}} = k,$$ where k is a the constant of proportionality. This can be further simplified, as follows: $$\frac{\sqrt{\frac{L}{g_1}}}{\sqrt{\frac{L}{g_2}}} = k$$ $$\sqrt{\frac{L}{g_1} \frac{g_2}{L}}= k$$ Whence $$\sqrt{\frac{g_2}{g_1}}= k.$$ To be able to answer, however, we must also recall from classical mechanics, specifically gravitation, that given two bodies, separated by a distance $$h$$, whose masses are $$M$$ and $$m$$, the gravitational force between them is given by: $$F_g = \frac{G M m}{h^2}.$$ Let $$M$$ be the mass of the Earth and $$m$$ the mass of the pendulum. Since Newton’s second law of motion dictates that force equals mass times acceleration, the ratio of the forces exerted on the pendulum on the different heights will be the same as the ratio between the accelerations, as long as the mass doesn’t change (which was a given of the problem). Therefore, we may write $$\sqrt{\frac{ \frac{G M m}{h_2^2}} {\frac{G M m}{h_1^2}} } = k$$ which, after simplification yields: $$\sqrt{\frac{h_2^2}{h_1^2}}=\frac{h_2}{h_1} = k$$ Hence, we conclude that $$\frac{T_1}{T_2}=\frac{h_2}{h_1} = k$$ Since we were asked for the change in frequency, we must remember that $$f=\frac{1}{T}$$, and thus $$\frac{f_2}{f_1}=\frac{T_1}{T_2}=\frac{h_2}{h_1} = k$$ so $$f_2 = \frac{h_2}{h_1} f_1$$ and the variation in frequency can be written as $$\Delta f = f_2-f_1 = \frac{h_2}{h_1} f_1 - f_1$$ which is what we were asked to calculate, so we are done.

## Contact tutor

Send a message explaining your
needs and Pedro will reply soon.
Contact Pedro

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage