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gavmur [86]
2 years ago
10

The resistivity of a given wire of cross sectional area 0.7mm is 4.9*10-4mm.Calculate the resistance of 2m length of the wire

Physics
1 answer:
Masja [62]2 years ago
3 0

Answer: 1.4\ \Omega

Explanation:

Given

Cross-sectional area A=0.7\ mm^2

Resistivity \rho =4.94\times 10^{-4}\ \Omega-mm

Length of wire l=2\ m\ or\ 2000\ mm

Resistance is given by

\Rightarrow R=\dfrac{\rho l}{A}\\\\\text{Insert values}\\\\\Rightarrow R=4.9\times 10^{-4}\times \dfrac{2000}{0.7}\\\\\Rightarrow R=1.4\ \Omega

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Nataliya [291]

the answer is definitely A.

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3 years ago
A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0
2 years ago
A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case
vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction,\rm \mu_k=0.520

Coefficient of static friction,\rm  \mu_s=0.520

The friction force, F=?

Resolve the force in the inclined plane;

\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times  cos 25 ^0 \\\\ F= 46.2 \ N

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

#SPJ1

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2 years ago
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That type of bending is called "diffraction" of waves.
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