Allison B.

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Physics (Newtonian Mechanics)

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Question:

You push a box of mass $$2\ \rm kg$$ (starting from rest) along a frictionless surface with a force of $$10\ \rm N$$ for $$5$$ seconds. The box then slides without forces acting on it along this surface for $$10\ \rm s$$, at which point it encounters a friction force of $$10\ \rm N$$ and slides until it stops. How far (in meters) has the block gone?

Allison B.

Answer:

We can divide this problem into three parts. First, the box is pushed with no friction. Second, the box glides with no friction. Third, the box slides as friction slows it down. Let's approach each of these parts as a miniature problem. $$\textbf{Part 1}$$: We want to find a distance, and we are given a mass, a force, a time, and an initial velocity (the box starts from rest, so $$v_0=0\ \rm m/s$$. We don't have an acceleration, but we can find one with Newton's Second Law. Therefore, we should use $$x=x_0+v_0t+\frac{1}{2}at^2$$. Plugging in $$a=\frac{F}{m}$$: $$x=0\ \rm m+(25\ \rm m/s)(5\ \rm s)+\frac{1}{2}(\frac{10\ \rm N}{2\ \rm kg})(5\ \rm s)^2$$ $$x=62.5\ \rm m$$ Now we have the distance traveled during the first part of the problem. We can use this as $$x_0$$ in the next part and calculate a cumulative distance traveled. $$\textbf{Part 2}$$: We know that no forces are acting on the box at this point, so according to Newton's Second Law, $$a=0\ \rm m/s^2$$. However, in order to use the equation we used for Part 1, we need the initial velocity. In order to find the speed at the end of Part 1, we can use $$v=v_0+at$$. Remember, here $$a$$ is the acceleration during Part 1, since we are trying to find the speed at the end of Part 1 (and consequently, the initial speed for Part 2). We will again use $$a=\frac{F}{m}$$. $$v=0\ \rm m/s+(\frac{10\ \rm N}{2\ \rm kg})(5\ \rm s)=25\ \rm m/s$$. Now we have everything we need to use $$x=x_0+v_0t+\frac{1}{2}at^2$$. Remember, for Part 2, $$a=0$$ and $$x_0=62.5\ \rm m$$ (the distance from Part 1). $$x=62.5\ \rm m+(25\ \rm m/s)(10\ \rm s)+0$$ $$x=312.5\ \rm m$$ $$\textbf{Part 3}$$: Finally, we reach the part where the box is slowing down under the influence of friction. We have to be careful this time, because the friction force is acting opposite to the direction in which the box is moving. This means that the acceleration resulting from this friction force is $$\textit{negative}$$. Before we can use our equation for $$x$$, however, we have to figure out how long it takes for the box to come to a rest. As in Part 2, we can use $$v=v_0+at$$. Since we want to find the time for the box to come to a stop, $$v=0\ \rm m/s$$. $$v_0$$ is going to be the final velocity from Part 2. Since the box was not accelerated in Part 2, the velocity stayed the same and therefore $$v_0=25\ \rm m/s$$. Our acceleration, again, is negative and comes from Newton's Second Law: $$a=\frac{-10\ \rm N}{2\ \rm kg}$$. $$0\ \rm m/s=25\ \rm m/s+(\frac{-10\ \rm N}{2\ \rm kg})t$$ $$t=5\ \rm s$$ Now, we can plug in one last time to find the cumulative distance traveled by the box. $$x=312.5\ \rm m+(25\ \rm m/s)(5\ \rm s)+\frac{1}{2}(\frac{-10\ \rm N}{2\ \rm kg})(5\ \rm s)^2$$ $$x=375\ \rm m$$ The box traveled $$\textbf{375 m}$$.

Pre-Calculus

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Question:

What are the asymptotes (vertical, horizontal, and/or slant) of the following graph? $$y=\frac{x^3+x^2-2x}{x^2-1}$$

Allison B.

Answer:

Our first step is always to factor the numerator and denominator completely. The denominator will tell us about any vertical asymptotes, and then we can look at the degree of the numerator and denominator to determine the presence of horizontal or slant asymptotes. It's important to do this step after we factor, because we may be able to simplify a bit. Let's look at the numerator first. There is a greatest common factor of x, so we take that out. $$x(x^2+x-2)$$ Now we have a quadratic, which we can factor in the usual way. We need a pair of numbers that multiplies to $$-2$$ and adds to $$1$$. These are $$2$$ and $$-1$$. So the numerator becomes $$x(x+2)(x-1)$$. Now for the denominator. We already have a quadratic here, and we can use DOTS because $$x^2$$ and $$1$$ are both perfect squares. The denominator is then $$(x-1)(x+1)$$. Notice that we have a factor of $$(x-1)$$ in both the numerator and the denominator. We can cancel those, which would give us a "hole" at the point where $$x=1$$. Our new function is therefore $$y=\frac{x(x+2)}{x+1}$$. To find the vertical asymptote, we set the denominator equal to zero and find $$x=-1$$. Notice that the numerator has a degree of 2 (if you multiply the factors back together, you get $$x^2$$ as the leading order), while the denominator has a degree of 1. Since the numerator is one higher than the denominator, we have a slant asymptote. To find its equation, we use synthetic division to divide the numerator by the denominator. In order to make this easier for ourselves, let's multiply the two remaining factors in the numerator back together to get $$x^2+2x$$. Notice that if we multiply $$(x+1)$$ by $$x$$, we get $$x^2+x$$, which is close to the expression in the numerator. So the first term in our slant asymptote is $$x$$. This leaves a remainder of $$x$$, which can be taken care of if we multiply our $$(x+1)$$ by 1. Now we have a remainder of 1, which we don't care about for the purposes of the slant asymptote. We're done! Our graph has a slant asymptote of $$y=x+1$$ with a hole at $$x=1$$, and a vertical asymptote of $$x=-1$$.

Astronomy

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Question:

Find the kinetic energy of a $$10\ \rm km$$-diameter asteroid with a bulk density of $$75\ \rm kg/m^3$$ that is approaching Earth at $$18\ \rm km/s$$. You may treat the asteroid as a sphere.

Allison B.

Answer:

We know that the formula for kinetic energy is $$K=\frac{1}{2}mv^2$$. However, we don't know what $$m$$ is. We are given a diameter and a density, so we can use this to find the mass via $$m=\rho V=\rho(\frac{4}{3}\pi r^3)$$. If the diameter is $$10\ \rm km$$, then the radius is $$5\ \rm km$$, or $$5\times10^3\ \rm m$$. So, plugging in: $$K=\frac{1}{2}\rho(\frac{4}{3}\pi r^3)v^2$$. Now that we have an equation in terms of things we are given, we can plug in our numbers. Here I'm going to write the equation after multiplying the $$\frac{1}{2}$$ and the $$\frac{4}{3}$$, just to simplify things a bit. Notice that since we are using meters, kilograms, and seconds, I've converted $$18\ \rm km/s$$ to $$1.8\times10^4\ \rm m/s$$. $$K=\frac{2}{3}(75\ \rm kg/m^3)\pi(5\times10^3\ \rm m)^3(1.8\times10^4\ \rm m/s)^2$$ $$K=6.4\times10^{21}\ \rm Joules$$

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