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3 years ago

A beam of red light and a beam of blue light have exactly the same energy. Which beam contains the greater number of photons?

1 answer:

4 0

The higher the frequency, the more energy the photon has. Of course, a beam of light has many photons. This means that really intense red light (lots of photons, with slightly lower energy)

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Please help!! The graph in the figure shows the position of a particle as it travels along the x-axis. What is the magnitude of

Bogdan [553]

Answer:

1.3 m/s

Explanation:

average speed = total distance/ total time

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3 years ago

What happens when tectonic plates move away from each other? Molten matter cools and sinks towards the core. Lava travels away f

satela [25.4K]

Molten Mattet Seeps through the crust and forms new land.

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3 years ago

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

blsea [12.9K]

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

$\rho (r) = \alpha (1-\frac{r}{R})$

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

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

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

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.

4 0

3 years ago

Consider a thick cylindrical wire of radius a. What must be the functional form of the current density J so that the magnetic fi

NikAS [45]

Answer:

Explanation:

Check the attachment

3 0

3 years ago

Nine-year-old Dakoda realizes that the quantity of water in a glass remains the same, even when the water is poured into a diffe

Archy [21]

Answer:

Conservation

Explanation:

She has observation conservation because If the temperature of the liquids stays constant and the container is insulated and not heat or cool the liquid much would not change the density of the liquid very much so that it's original volume could remain constant.

The interesting thing is not that the child assumes the taller glass holds more liquid but that they fail to understand conservation: the fact that the water from one glass is going to be the same amount after being poured into any other container. It's as if they did not realize the water came from the same glass.

3 0

3 years ago