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Sloan [31]
3 years ago
15

You take a 3 hour nap on a train traveling 40 miles per hour

Physics
2 answers:
Ganezh [65]3 years ago
7 0
120 miles - 3 x 40 = 120
Svetradugi [14.3K]3 years ago
5 0

Answer:

120 miles

Explanation:

Speed = distance traveled/ time taken.

Distance traveled = speed * time taken

Speed = 40 miles per hour, time = 3 hours.

Distance = 40*3

The train's traveled distance is 120 miles.

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A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut
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Answer:

1.E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2.E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

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We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

\int\, da = 2\pi r h

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Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3. At the boundary where r = R:

E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}

As can be seen from above, two E-field values are equal as predicted.

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