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Tutor profile: Dima W.

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Dima W.
BS In Astronomy/Astrophysics, Extragalactic Research, SMe Research
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Questions

Subject:Calculus

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Question:

Why is Calculus renown for being so difficult?

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Dima W.

Calculus is renown for being difficult because, in order to understand the subject matter presented in Calculus, it requires confidence in areas of mathematics that you (should have) studied earlier such as algebra and geometry. In particular, those who are not entirely familiar with the rules of algebra will find difficulty with calculus because calculus requires much knowledge in the basic laws of algebra. If you're using James Stewart's "Calculus: Early Transcendentals" textbook, be sure to read the first chapter of the textbook for a quick review of all the kinds of mathematical tools and techniques that will show up later in the book. Calculus also asks you to be comfortable with abstraction. For example, can you imagine, if you "zoomed-in enough" on a point on the curve $$y = x^{2}$$, that the curve actually "flattens out" into a straight line? These ideas are important in defining operations such as limits or derivatives. Despite its notoriety, as with all forms of mathematics, calculus comes with practice. I recommend you to not be discouraged by getting problems wrong or by those who "seem to know it all." Furthermore, because calculus has gained such a "scary" reputation lately, there are now many online resources to help students struggling with the subject. There is a lot of help out there.

Subject:Astronomy

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Question:

What actually is "magnitude" in astronomy?

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Dima W.

A "magnitude" is generally a measure of the brightness of an object. There are two "types" of magnitudes that come up often: the "apparent" magnitude and the "absolute" magnitude. The apparent magnitude is self-descriptive: it is the apparent brightness of an object. It is essentially eyeballing an object and asking how bright it is. Which astronomical object do you think has the largest apparent magnitude? Don't think too hard. The absolute magnitude is a measure of the intrinsic brightness of an object. We know our Sun is bright, but imagine if we replaced it with one of those other stars far away. Would that new star be brighter or dimmer than the Sun? That would depend on the absolute magnitude of those stars. Unfortunately, because of convention, the more negative a magnitude is, the brighter the object is. The Sun's apparent magnitude of $$-26.7$$ actually means it's very bright. Compare that with a hypothetical, distant star whose $$\textit{absolute}$$ magnitude might actually be larger than our Sun's, but has an $$\textit{apparent}$$ magnitude of $$+15$$.

Subject:Physics

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Question:

What is $$PE = mgh$$ and how do I use it?

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Dima W.

Potential energy ($$PE$$) is unfortunately one of those non-intuitive physics things that only more makes sense the more you use it. $$PE = mgh$$ is the formula for the gravitational potential energy of an object a height $h$ above the ground. $$PE$$ is just a number. If an object of mass $$m$$ is suspended some height $$h$$ above the ground, it will have a $$PE$$ of $$mgh$$. If the object is lying on the ground, is there any $$\textit{potential}$$ for it to move? In that case, $$h = 0$$, so $$PE = 0$$. This formulation of $$PE$$ is used when considering objects that are "close" to the surface of the Earth. The constant $$g$$ is the acceleration due to gravity of an object "close" to the surface of Earth (roughly $$9.81 {m}/{s^{2}}$$.) Usually, this form of $$PE$$ is used in asking and solving questions such as "what is the final speed of an object of mass $$m$$ released from rest at a height $$h$$ above the surface of the Earth?" All you have to do is use conservation of energy, $$E_{initial} = E_{final}$$, where the form of $$PE$$ in each $$E = TE + PE$$ is $$PE = mgh$$.

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