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2 years ago

18) A charm quark has a charge of approximately

Physics

1 answer:

Allushta [10]2 years ago

3 0

Based on scientific records, a charm quark has a charge that's approximately equal to: 2) 1.07 × 10⁻¹⁹ C.

<h3>What is a charge?</h3>

A charge simply refers to a fundamental, physical property of matter that governs how the particles of a substance are affected by an electromagnetic field, especially due to the presence of an electrostatic force (F).

Also, charge is typically measured in Coulombs and a charm quark (elementary particle) has a charge that's approximately equal to 1.07 × 10⁻¹⁹ Coulomb.

Read more on charges here: brainly.com/question/4313738

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The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig

miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

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3 years ago

When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.

Ivan

When an object in simple harmonic motion is at its maximum displacement, its <u>acceleration</u> is also at a maximum.

<u><em>Reason</em></u><em>: The speed is zero when the simple harmonic motion is at its maximum displacement, however, the acceleration is the rate of change of velocity. The velocity reverses the direction at that point therefore its rate of change is maximum at that moment. thus the acceleration is at its maximum at this point</em>

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2 years ago

Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium

ivolga24 [154]

Answer:

Explanation:

Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

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3 years ago

What is sedentary lifestyle?

Effectus [21]

A sedentary lifestyle is when you don't make any exercise. For example, if I spent a whole week just sitting down answering questions, that would be an example of a sedentary lifestyle, because I'm not running or jogging. Having that type of lifestyle can make you fat and get health problems.

Have a nice day! :)

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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