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4 years ago

What medium is often used to transfer computer information?

Physics

2 answers:

Anika [276]4 years ago

6 0

Optic fiber is the medium that is often used to transfer computer information

dsp734 years ago

3 0

The correct answer is D.

https://en.wikipedia.org/wiki/Optical_fiber<span>
Optical fibers are used most often as a means to transmit light between the two ends of the fiber and find wide usage in fiber-optic communications, where they permit transmission over longer distances and at higher bandwidths (data rates) than wire cables.</span>

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Someone please help! It is a few science questions. Fairly easy!

borishaifa [10]

1) $0.92 * 100 = 92$ <Solve using the formula which is:
Mass=Density×Volume
$M=D*V$

2) $3.98 ml$

3) $V=L*W*H$ (Length × Width × Height)

$5*2*4=40$
$\frac{300}{40} =7.5$ (Answer=7.5)

3 0

4 years ago

Read 2 more answers

A thin slab of Germanium is used as a Hall Effect probe. How would you orient a magnetic field to make the side facing out of th

Cerrena [4.2K]

Answer:

the magnetic field must go in a direction parallel to the page perpendicular to the current.

Explanation:

The Hall effect is the voltage produced by the movement of electrons due to the effect of electric and magnetic fields in a material

F = eE + v x B

The electric field goes in the direction of the current that is opposite to the direction of the electrons, therefore the magnetic force must be perpendicular to it.

Therefore, if the current goes in a direction parallel to the page, in the x direction, the magnetic field must be perpendicular to it if we use the rule of the right wizard,

thumb points in the direction of E, x axis parallel page

The fingers extended should go parallel to the page in the direction and up

The palm is the direction of the Force, where the voltage will be produced points out the page, this is for positive charges, as in germanium the charges are negative, the real force goes into the page.

Therefore the electrons accumulate on the inside of the page and the voltage is negative in this part.

Therefore the voltage is positive on the outside of the sheet. In conclusion the magnetic field must go in a direction parallel to the page perpendicular to the current.

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3 years ago

how far from the lens is the image of the house if the house is 16 ft from the thin convex lens with a focal length of 8 ft

BlackZzzverrR [31]

This question involves the concepts of the thin lens formula, focal length, and image distance.

The image of the house is "16 ft" away from the lens.

According to the thin lens formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where,

f = focal length = 8 ft

p = object distance = 16 ft

q = image distance = ?

Therefore,

$\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\$

Learn more about the thin lens formula here:

brainly.com/question/3074650

6 0

3 years ago

During blank he is transferred by the movement of currents within a fluid

skelet666 [1.2K]

You need to put more info into the question

6 0

3 years ago

Uranium was used to generate electricity in a nuclear reactor of a nuclear power station. The nuclear reactor was 29% efficient

Minchanka [31]

Answer:

$C_{lifetime} = 2,618,017,174\,USD$

Explanation:

The energy contained per second in the fuel is:

$\dot E_{fuel} = \frac{800\times 10^{6}\,W}{0.29}$

$\dot E_{fuel} = 2.758\times 10^{9}\,W$

It is know that fission of a gram of uranium liberates $5.8\times 10^{8}\,J$ . Mass flow rate is computed herein:

$\dot m_{U}= \frac{2.758\times 10^{9}\,W}{5.8\times 10^{8}\,\frac{J}{g}}$

$\dot m_{U} = 4.755\,\frac{g}{s}$

The needed quantity of fuel per second is:

$\dot m_{fuel} = \frac{4.755\,\frac{g}{s} }{0.04}$

$\dot m_{fuel} = 118.875\,\frac{g}{s}$

The annual fuel consumption is now obtained:

$\Delta m_{fuel,year} = (0.119\,\frac{kg}{s})\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{24\,h}{1\,day} )\cdot(\frac{365\,days}{1\,year} )$

$\Delta m_{fuel,year} = 3,752,784\,\frac{kg}{year}$

The cost of fuel for the lifetime of the plant is:

$C_{lifetime} = (3,752,784\,\frac{kg}{year} )\cdot (\frac{0.453\,lb}{1\,kg} )\cdot [(20\,years)\cdot(\frac{12\,USD}{1\,lb} )+(20\,years)\cdot (\frac{65\,USD}{1\,lb} )$

$C_{lifetime} = 2,618,017,174\,USD$

6 0

3 years ago