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mel-nik [20]
2 years ago
7

Between 1992 and today, astronomers using large telescopes have discovered many icy pieces that orbit in the same region as the

orbit of Pluto. These are believed to be members of the
Physics
1 answer:
pantera1 [17]2 years ago
6 0

Icy pieces that orbit in the same region as the orbit of Pluto are believed to be members of the Kuiper belt.

<h3>What is the Kuiper belt?</h3>

The Kuiper belt is a shape-disc formation in the solar system observed from Neptune and more away from our sun.

The Kuiper belt consists of different celestial bodies including among others icy pieces, planets, and comets.

In conclusion, icy pieces that orbit in the same region as the orbit of Pluto are believed to be members of the Kuiper belt.

Learn more about the Kuiper belt here:

brainly.com/question/3955749

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Which of the following situations describes a change that will result in a new kind of matter with different characteristics?
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Chemical property - characteristic of something that allows it to change to something new.

Explanation:

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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

Total charge, Q = 7 nC = 7\times 10^{- 9} C

At x = 2L,

Electric field, \vec{E_{2L}} = 500N/C

Coulomb constant, K = 8.99\times 10^{9} N.m^{2}/C^{2}

Now, we know that:

\vec{E} = K\frac{Q}{x^{2}}

Also the line charge density:

\lambda = \frac{Q}{L}

Thus

Q = \lambda L

Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

4 0
4 years ago
What is Snell law of refraction????​
Morgarella [4.7K]

\huge\mathbb\colorbox{black}{\color{white}{⏫FORMULA⏫}}

\huge\red {\underline { \overline{ \bf \mid\: \:  \: ↓ANSWER↓ \: \: \mid}}}

Explanation:

Snell's law (also known as Snell–Descartes law and the law of refraction) is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.

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