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nikitadnepr [17]
3 years ago
9

Soils, forests, and minerals are among earth’s _____ resources.

Physics
2 answers:
ddd [48]3 years ago
8 0

Soils, forests, and minerals are among Earth’s _____ resources.

LAND Resources! this was the correct answer for me

stich3 [128]3 years ago
7 0
Soils, forests and minerals are among earth's "Natural Resources".
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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
Which property of gold allows it to be used this way?
densk [106]

the electric conductivity of gold is very high

3 0
3 years ago
Which statement is true of a glass lens that diverges light in air?
Arisa [49]
C it is uniformly thick
3 0
2 years ago
Read 2 more answers
Match the object with its characteristic.
Triss [41]
What object do you need to match
8 0
3 years ago
Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te
zlopas [31]

Answer: C

Explanation: Side post terminals need to be removed to inspect them for corrosion.

Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.

4 0
3 years ago
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