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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

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

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

_____ are pictures of relationships

butalik [34]

The blank in the question can be filled with the word, “Graph”. Therefore, Graphs are the pictures which are in relationships.

<u>Explanation: </u>

Graph usually represents a set of data which is nonlinear in occurrence and has some relationship between the two given data. And as graph are pictorial representation, it is simply assumed as the pictures of relationships.

For example, a graph can be drawn for the set of data for the presence of number of students of all the sections of the particular class of a school, as they are relative. But making the graph for number of students in all section of all class but different school cannot be done as non-relative.

4 0

3 years ago

Read 2 more answers

If there is no slipping, a frictional force must exist between the wheels and the ground. in what direction does the frictional

djyliett [7]

The frictional force is always act in opposite direction of motion. if wheels is moving in +x axis or right then the f force is act in -x or left

8 0

2 years ago

The length of daylight on the moon is about __________.

Lyrx [107]

<span>The length of daylight on the moon is about 29.5 days.</span>

From the earliest days, the Moon has been there in the Solar System and there has never been a period when we couldn't gaze upward in the night sky and either observe the Moon hanging there, or realize that it would be back the precise one night from now (i.e. a New Moon).

A day on the Moon keeps going as long as 29.5 Earth days. We can say that it would take 29.5 days for the Sun to move the distance over the sky and come back to its unique position once more.

3 0

3 years ago

Read 2 more answers

The 25 kg wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it

N76 [4]

Answer:

Explanation:

radius of gyration of wheel k then

k² = r²/2

r² = 2 k²

r = √2 k

= 1.414 x .3 m

r = .4242 m

Moment of inertia of wheel

= mass x radius of gyration ²

= 25 x .3 x .3

= 2.25 kg m²

Friction force acting on it ( sliding )

= μmg , μ being coefficient of kinetic friction

This friction force will create linear acceleration in forward direction

Acceleration produced

= μg

= .6 x 9.8

= 5.88 m / s ²

This will also rotate the wheel , angular acceleration being

linear acceleration / radius

= 5.88 /.4242

= 13.86 radian / s²

3 0

2 years ago