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4 years ago

10

A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 280 Ω at room temperature, what is

the cross-sectional radius of the bar?

1 answer:

monitta4 years ago

4 0

Answer:

r = 17.05 cm

Explanation:

Given that,

Length of silicon bar is 4 cm or 0.04 m

Resistance of the bar is 280 ohms

We know that the resistivity of the silicon is 640 Ωm

We need to find the cross-sectional radius of the bar. Let it is r.

Using definition of resistance of an object. It is given by :

$R=\rho\dfrac{l}{A}$

A is area of bar, A = πr²

So,

$R=\rho\dfrac{l}{\pi r^2}\\\\r^2=\dfrac{\rho l}{R\pi}\\\\r^2=\dfrac{640\times 0.04}{280\pi}\\\\r=0.1705\ m\\\\r=17.05\ cm$

So, the cross-sectional radius of the bar is 17.05 cm.

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The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

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Let,

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Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

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An ore car of mass 35000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 17 m lower vert

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Answer:

Explanation:

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A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move

DochEvi [55]

Answer:

$W=-2.1405\times 10^9\,J$

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charge, $Q=-13\,\mu C=-13\times 10^{-6}\,C$

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$\theta= 47.75^{\circ}$

Now from the relation between the change in potential difference:

$\Delta V= E.d.cos\,\theta$

$\Delta V= 148\times 279.63\times cos\,47.75^{\circ}$

$\Delta V= 27826.06 V$

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∴ $\Delta V=\frac{W}{Q}$

$W=\frac{\Delta V}{Q}$

Putting the respective values

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$W=-2.1405\times 10^9\,J$

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