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ankoles [38]
3 years ago
7

What are the steps to extract metal from the earths crust

Physics
1 answer:
Pani-rosa [81]3 years ago
4 0
<span>Methods of extraction include: extract by electrolysis, extract by reaction with carbon or carbon monoxide, and extracted by various chemical reactions.</span>
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light incident on a polarizer is then passed through a second polarizer. if the polarizer and the analyzer are perpendicular to
Lerok [7]
<span>Indeed, this is one of the odd results in physics. A system of two polarizing filters arranged as shown below trasmits no light.</span>
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3 years ago
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There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
taurus [48]

Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

The electric potential at a distance d due to 'Q' is:

V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

V_{total} = \frac{q}{4\pi\epsilon_{o}d}

Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0
3 years ago
Please help on this one?
koban [17]

Using the given equation:

di = 20.0 * 10.0 / 20.0 - 10.0

di = 200/10

di = 20.0 cm

The answer is A.

6 0
3 years ago
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A bulb pile is driven to the ground with a 2.5 ton hammer. The drop height is 22 ft and the volume in last batch driven is 4 cu
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Answer:

159.1 ton

Explanation:

The solution is shown in the attached file

Download docx
5 0
3 years ago
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Help me with this review question please.
QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

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