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3 years ago

13

Suppose that an electric charge is produced on one part of a body. If the charge spreads through the entire body, the body is mo

st likely made of A. porcelain. B. plastic. C. metal. D. oil.

2 answers:

timurjin [86]3 years ago

8 0

C metal. most metals are a conductor which an electric current is attracted to.

spin [16.1K]3 years ago

8 0

C. Metal as it conducts

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A plan to budget time for studying and activities is referred to as? a study routine study habits study skills a study schedule

ad-work [718]

Answer:

a study schedule

Explanation:

5 0

3 years ago

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

The index of refraction for red light in water is 1.331 and that for blue light is 1.340. If a ray of white light enters the wat

kobusy [5.1K]

Answer:

Angle of refraction for red light is $43.01^{\circ}$

Angle of refraction for blue light is $42.68^{\circ}$

Explanation:

It is given refractive index for red light is $\mu _{red}=1.331$

Refractive index of blue light $\mu _{blue}=1.340$

Angle of incidence $i=65.30^{\circ}$

According to law of refraction $\mu =\frac{sini}{sinr}$

For red light $1.331 =\frac{sin65.30^{\circ}}{sinr}$

$1.331 =\frac{0.908}{sinr}$

$sinr=0.682$

$r=43.01^{\circ}$

Therefore angle of refraction for red light is $43.01^{\circ}$

Similarly for blue light $1.340 =\frac{sin65.30^{\circ}}{sinr}$

$1.340 =\frac{0.908}{sinr}$

$sinr=0.677$

r = $42.68^{\circ}$

Therefore angle of refraction for blue light is $42.68^{\circ}$

6 0

3 years ago

Could an average star, such as our sun, become a neutron star?

algol [13]

No but the sun could be a white dwarf stellar remnant.

3 0

4 years ago

In which direction are the winds blowing in the northern and southern hemispheres

Alika [10]

The winds normally blow from east to the west rather than blowing from the north to the south in the northern and the southern hemispheres.

<u>Explanation:</u>

The reason for the blowing of the wind from the east to the west rather than from the north to the south in both the hemispheres which are the northern and the southern hemispheres are because it happens in light of the fact that Earth's rotation around it's own axis creates what is known as the Coriolis effect.

The Coriolis impact makes wind frameworks turn counter-clockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere.

6 0

3 years ago