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# Tutor profile: Taylor P.

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Taylor P.
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## Questions

### Subject:Physics (Electricity and Magnetism)

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Question:

A solenoid that is $$95.0 cm$$ long has a radius of $$2.00 cm$$ and a winding of $$1200$$ turns; it carries a current of $$3.60 A$$. Calculate the magnitude of the magnetic field inside the solenoid.

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Taylor P.

Based on experimental observations, we know that moving charges generate magnetic fields. A solenoid is a wire configuration that resembles a slinky: wire is coiled into a cylindrical shape. since we know that moving charges generate a magnetic field around the wire, if we consider a solenoid, we find that the magnetic field inside the solenoid is a constant and oriented through the coils from top to bottom. Using Ampere's Law, we can prove that the magnetic field inside a solenoid has a magnitude of $( B = \mu_0in$) Where $$\mu_0$$ is the permeability of free space ( $$\mu_0 = 4\pi*10^{-7} N/A^2$$ ), $$i$$ is the current in the wire in Amps, and $$n$$ is the number of turns per unit length. In our particular problem, the current is given as $$i= 3.60A$$ and the turn densidty can be found using $(n=\frac{N}{l}=\frac{1200}{.950m}=1263.2 \frac{turns}{m}$) Therefore, we can calculate the magnitude of the magnetic field as $(B=(4\pi*10^{-7}N/A^2)(3.60A)(1263.2 \frac{turns}{m})=5.71mT$)

### Subject:Calculus

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Question:

Evaluate the integral: $( \int ln(x^2-x+2)dx$)

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Taylor P.

The integral is evidently not a simple problem we know from basic integration techniques (polynomials, trig functions, exponential, etc.), so we should begin evaluating which integration techniques would be useful for this problem. $(\\$) The first technique we typically consider is u-substitution. In this case, u-sub would likely not be useful because the only obvious choice for the 'u' is $$x^2-x+2$$, but then the 'du' would be $$2x-1$$, which don't have elsewhere in the integrand to use. $(\\$) The next technique we should try is integration by parts. In this case, we might try to assign variables as follows: $(u= ln(x^2-x+2) \hspace{0.4in} du = \frac{2x-1}{x^2-x+2} \\ dv = dx \hspace{0.4in} v = x$) Following the integration by parts method, we find $( \int ln(x^2-x+2)dx = xln(x^2-x+2)-\int \frac{2x^2-x}{x^2-x+2}dx$) The new integral may now be solvable through a combination of polynomial division and partial fraction decomposition. Dividing the two fractions and factoring the denominator, we obtain $( 2+\frac{x-4}{(x-2)(x+1)}$) Applying partial fractions decomposition, we get $(2-\frac{\frac{2}{3}}{x-2}+\frac{\frac{5}{3}}{x+1}$) Integrating, we find the final solution $(\int ln(x^2-x+2)dx = xln(x^2-x+2)-2x+ \frac{2}{3}ln|x-2| - \frac{5}{3} ln|x+1|$)

### Subject:Physics

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Question:

If a source is moving towards a detector at $$55$$ $$m/s$$ while the detector is moving towards the source at $$25$$ $$m/s$$, what would the detected frequency be for a source emitting sound waves at $$500$$ $$Hz$$?

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Taylor P.

This question relies on the Doppler effect. The Doppler effect is similar to the difference in pitch between an ambulance that is approaching you versus one that is driving away. The Doppler effect is governed by the equation: $( f' = f \frac{v \pm v_D}{v \pm v_S}$) Where $$f'$$ is the detected frequency, $$f$$ is the emitted frequency, $$v$$ is the speed of sound in air (typically given as $$343$$ $$m/s$$), $$v_D$$ is the speed of the detector, and $$v_S$$ is the speed of the source. $(\\$) The next step is to determine whether to add or subtract the speeds in the numerator and denominator for this Doppler effect formula. the key to doing this is to determine whether the motion of the source or detector would $$\textit{increase}$$ or $$\textit{decrease}$$ the detected frequency. In this case, since the source is approaching the detector, we expect the source's motion to $$\textit{increase}$$ the detected frequency, thus we must $$\textit{subtract}$$ the speed of sound and the speed of the source in the denominator. Additionally, since the detector is approaching the source, we expect this to $$\textit{increase}$$ the detected frequency, thus we $$\textit{add}$$ the speed of sound and speed of the detector in the numerator. $(\\$) Therefore, the solution to this problem would be $( f' = 500 Hz \frac{343 m/s + 25 m/s}{343 m/s - 55 m/s} = 638.9 Hz$)

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