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3 years ago

You take a conductive wire and make loops in it, you create a _________.

1 answer:

7 0

When you take a long piece of wire, form it into a coil of several loops, and run a current through it, you get a magnetic field that runs straight along the axis of the coil.

This arrangement of a conductor and the current through it is called a "<em>solenoid</em>". <em>(C) </em>

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Directions: Answer the following True or False questions using a pencil.

anygoal [31]

Answer:

true

Explanation:

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3 years ago

Which factor affects the amount of runoff that occurs in an area?

jarptica [38.1K]

ion:

h t t p s : / / w w w . y o u t u b e . c o m / w a t c h ? v = x a a z U g E K u V A

this is the video that explains it

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4 years ago

An electromagnet is strong enough to lift 200 kg of mass. If you double the number of turns of wire around the core and triple t

jekas [21]

Magnetic field is proportional to NIA.

N is the number of turns, I is the current and A is the area

You see N=2n, I = 3 i. Therefore, it will 6 times stronger

200*6= 1200 kg. It can lift now is 1200 kg.

6 0

3 years ago

Read 2 more answers

Calculate the current passing through a conductor of resistance 4ohms. If a potential difference of 15V its ends______

statuscvo [17]

Explanation:

current = velocity/resistance

I = V/R

15/4

current = 3.75A

hope this helps...

6 0

3 years ago

(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. Th

KonstantinChe [14]

Answer:

Part A:

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=3.03*10^{-3} m^2/V$

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

$N-Au=\frac{Density*Avogadro Number}{atomic weight}$

$Density = 19.32g/cm^3$

$Avogadro Number=6.02*10^{23} atoms/mol$

$Atomic weight=196.97g/mol$

So:

$n=1.5*\frac{Density*Avogadro Number}{atomic weight}$

$n=1.5*\frac{19.32*6.02*10^{23}}{196.97}$

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}$

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

$Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}$

$Electron Mobility=3.03*10^{-3} m^2/V$

4 0

4 years ago