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# Tutor profile: Aaron S.

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Aaron S.
Chemistry and physics teacher; 5 years tutoring experience
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## Questions

### Subject:Chemistry

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Question:

Sigmund has a chunk of metal and he want to determine what kind of metal it is. He knows that it is a pure metal, and he has access to a list of metals cross-referenced with their specific heats, so he decides to use calorimetry. Using a mass balance, Sigmund determines that the mass of his chunk of metal is 237 grams. He then boils the chunk for a lengthy period of time before dropping it into 500.0 mL of water at 20.00 $$^{\circ}$$C. If the temperature of the water rises to 28.94 $$^{\circ}$$C, what is the identity of his metal? (Assume a perfectly insulated calorimeter and an atmospheric pressure of 1.000 atm).

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Aaron S.

The key to any calorimetry problem is to remember that any heat energy lost by the metal must be picked up by the fluid. This allows us to set up an equation with the metal on one side and the water on the other. Remember that the equation for specific heat is: $(Q = cm\Delta T$)Where $$Q$$ is the heat added to the substance, $$c$$ is the specific heat capacity of the substance, $$m$$ is the mass of the sample, and $$\Delta T$$ is the change in temperature of the substance. We have already established that any heat that exits the metal must enter the water, or, in mathematical terms: $(-Q_{metal} = Q_{water}$)The heat for the metal is negative because energy is leaving the metal, while the heat for the water is positive because energy is entering the water. Looking back up at our first equation, we can substitute in $$cm\Delta T$$ for each $$Q$$:$(-c_{metal}m_{metal}\Delta T_{metal} = c_{water}m_{water}\Delta T_{water}$)From here we can plug in our numbers. Remember that since our metal was boiled it will enter the water at a temperature of 100.00 $$^{\circ}$$C:$(-c_{metal} \times 237 g \times (28.94^{\circ}C - 100.00^{\circ}C) = 4.186 J/gC^{\circ} \times 500.0g \times (28.94^{\circ}C - 20.00^{\circ}C)$) From here we can rearrange the equation by algebra and solve for $$c_{metal}$$:$(c_{metal} = -\frac{237 g \times (28.94^{\circ}C - 100.00^{\circ}C)}{4.186 J/gC^{\circ} \times 500.0g \times (28.94^{\circ}C - 20.00^{\circ}C)} = 0.900 J/gC^{\circ}$)Sigmund checks this result against a table of specific heat capacities of various metals and finds that this specific heat capacity corresponds to that of aluminum.

### Subject:Calculus

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Question:

Evaluate the definite integral $(\int_{0}^{3\sqrt3}\frac{x}{\sqrt{x^2 + 9}}dx$)

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Aaron S.

This problem is too complex to simply solve in one fell swoop, but it becomes much simpler if we use u-substitution. Choosing how to define our $$u$$ is the most important part of a problem like this. Whatever is under the radical is usually a good choice, so let's try that:$(u = x^2 + 9$)$(du = 2xdx$)First, let's substitute in our $$u$$ to see if it's going to work out:$(\int_{x=0}^{x=3\sqrt3}\frac{x}{\sqrt{u}}dx$)Notice that our integral has an $$xdx$$ in it. This is very similar to $$du = 2xdx$$. In fact, algebra tells us that: $(xdx =\frac{1}{2}du$)It looks like our choice will work out nicely. Let's substitute in our $$du$$:$(\int_{x=0}^{x=3\sqrt3}\frac{\frac{1}{2}du}{\sqrt{u}}$)$$\frac{1}{2}$$is a constant, so we can move that out in front of our integral. With a little rearranging we can turn our integral into this:$(\frac{1}{2}\int_{x=0}^{x=3\sqrt3}\frac{1}{\sqrt{u}}du$)Let's do some algebra:$(\frac{1}{\sqrt{u}} = \frac{1}{u^{\frac{1}{2}}} = u^{-\frac{1}{2}}$)So really, our integral is:$(\frac{1}{2}\int_{x=0}^{x=3\sqrt3}u^{-\frac{1}{2}}du$)Time to solve our integral. Remember this general equation for an integral:$(\int{x^{n}dx} = \frac{x^{n+1}}{n+1}+C$)In our equation $$x\rightarrow u$$ and $$n\rightarrow -\frac{1}{2}$$ (and since it's a definite integral we don't need to worry about the $$+C$$)$(\int{u^{-\frac{1}{2}}du} = \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = 2u^{\frac{1}{2}}$)With our integral solved, we can plug in our coefficients and parameters:$(\frac{1}{2}\int_{x=0}^{x=3\sqrt3}u^{-\frac{1}{2}}du = \frac{1}{2}(2u^{\frac{1}{2}})\Big|_{x=0}^{x=3\sqrt3} = u^{\frac{1}{2}}\Big|_{x=0}^{x=3\sqrt3}$)Finally, we can plug our $$u = x^2 + 9$$ back in:$(u^{\frac{1}{2}}\Big|_{x=0}^{x=3\sqrt3} = (x^2 + 9)^{\frac{1}{2}}\Big|_{0}^{3\sqrt3} = \sqrt{x^2 + 9}\Big|_{0}^{3\sqrt3}$)And solve:$(\sqrt{(3\sqrt3)^2 + 9} - \sqrt{0^2+9} = \sqrt{27 + 9} - \sqrt{9} = 6 - 3 = 3$)

### Subject:Physics

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Question:

A bullet (mass 35 g, velocity 200.0 m/s) collides with a pendulum (length 2.3 m, mass 1.665 kg), embedding itself in the bob. How high will the pendulum bob swing before stopping?

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Aaron S.

The first step in solving this problem is to recognize that since the bullet sticks in the pendulum bob this must be a perfectly inelastic collision. Momentum is always conserved in a collision, and since this collision is perfectly inelastic we know that at the end of the collision the bob and the bullet must be moving at the same velocity. Our equation for the collision, therefore, is: $(m_{bullet}v_{bullet} + m_{bob}v_{bob} = (m_{bullet} + m_{bob})v_{bullet + bob}$) Plugging in our numbers, we get: $(0.035 kg \times 200.0 m/s + 1.665 kg \times 0 m/s = (0.035 kg + 1.665 kg) \times v_{bullet + bob}$)Rearranging this problem by algebra and solving gives us: $(v_{bullet + bob} = \frac{0.035 kg \times 200.0 m/s + 1.665 kg \times 0 m/s}{0.035 kg + 1.665 kg} = 4.1176 m/s$)Now that we know how fast the (bullet + bob) is moving after the collision, we can use conservation of energy to determine how far up the system swings before stopping. Since we're dealing with motion and height, the only two forms of energy we need to worry about are translational kinetic energy and gravitational potential energy. Our equation for the conservation of energy, then, is: $(\frac{1}{2}m_iv_i^{2} + m_igh_i = \frac{1}{2}m_fv_f^{2} + m_fgh_f$)Since the system is at a height of 0 right after the collision it will start with 0 gravitational potential energy, and since the system will stop momentarily at the top of its swing, it will end with 0 translational kinetic energy. Since both of these terms will be 0, they can be dropped, and our equation will become: $(\frac{1}{2}m_iv_i^{2} = m_fgh_f$)Notice that there is a mass term on both sides of this equation. Since mass is fixed throughout this movement, we can cancel it, simplifying our equation to: $(\frac{1}{2}v_i^{2} = gh_f$)Plugging in numbers, we get: $(\frac{1}{2}(4.1176m/s)^{2} = 9.80m/s^2 \times h_f$)Rearranging this problem by algebra and solving gives us our final height: $(h_f = \frac{\frac{1}{2}(4.1176m/s)^{2}}{9.80m/s^2} = 0.865 m$)

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