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Natasha2012 [34]
3 years ago
8

A wire with a circular cross section and a resistance R is lengthened to 9.66 times its original length by pulling it through a

small hole. The total volume of the wire is unchanged. Find the resistance of the wire after it is stretched. Answer in units of R
Physics
1 answer:
damaskus [11]3 years ago
8 0

Answer:

The resistance of the wire after it is stretched is 93.31R.

Explanation:

Resistance is the property of the material to oppose the current flow through it. It is given by the relation :

R = (ρl)/A

Here ρ is resistivity, l is length of wire and A is the area of the wire.

Let l₀, and A₀ are the original length and original circular cross section area of the wire. while l₁ and A₁ are the new length and new circular cross section area of the wire.

Volume of the original wire, V₀ = A₀ x l₀

Volume of the new wire, V₁ = A₁ x l₁

According to the problem. volume remain same. So,

V₀ = V₁

A₀ x l₀ = A₁ x l₁

It is given that l₁ = 9.66 x l₀. Substitute this value in the above equation;

A₀ x l₀ = A₁ x 9.66 x l₀

A₁ = A₀/9.66

Resistance of the original wire, R = (ρl₀)/A₀

Resistance of the new wire, R₁ = (ρl₁)/A₁

Substitute the value of l₁ and A₁ in the above equation.

R₁ = (ρ x l₀ x 9.66)/(A₀/9.66) = 93.31 x (ρl₀)/A₀

But (ρl₀)/A₀ = R. hence,

R₁ = 93.31 R

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ozzi

Answer:

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In Aristotle's Metaphysics, there are four causes of existence or change in nature: the material cause, the formal cause, the efficient cause, and the final cause.

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3 0
3 years ago
Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.0677 V. How much work is done
Lina20 [59]

Answer: Recall that

Work (W) = - e × ∆V

Where

Charge of electron= 1.6 × 10^ -19 c

And from the question we were given change in potential as = -0.0677V

Work = - (1.6 × 10^-19) × (-0.0677)

Work= 1.0832 × 10^-20J

Therefore the work done by the electric force when a sodium ion (charge = +e) moves from the outside to the inside is 1.0832×10^-20

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3 years ago
An aluminum wire having a cross-sectional area equal to 3.00 10-6 m2 carries a current of 7.00 A. The density of aluminum is 2.7
Natali [406]

Answer:

Vd = 2.42 ×10⁻⁴ m/s

Explanation:

Given: A = 3.00×10⁻⁶ m², I = 7.00 A, ρ = 2.70 g/cm³

To find Drift Velocity Vd=?

Sol

the formula is Vd = I/nqA             (n is the number of charge per unit volume)

n = No. of electron in a mole ( Avogadro's No.) / Volume

Volume = Molar mass / density   ( molar mass of Al =27 g)

V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³

n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)

n= 6.02 × 10 ²⁸

Now

Vd = (7A) / (  6.02 × 10 ²⁸ ×  1.6 × 10⁻¹⁹ C ×  3.00×10⁻⁶ m²)

Vd = 2.42 ×10⁻⁴ m/s

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4 years ago
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude
bearhunter [10]

Answer:

c.

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A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const
avanturin [10]

Answer:

0.17547 m

Explanation:

m = Mass of block = 3.3\times 10^{-2}\ kg

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m

The amplitude of the resulting simple harmonic motion is 0.17547 m

6 0
3 years ago
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