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padilas [110]
3 years ago
7

A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho

= _____ Ω⋅m
Physics
1 answer:
spayn [35]3 years ago
4 0
<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A               ------------------------(i)

Where;

A = πd² / 4              [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4      

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

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Answer:

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

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*Effect of friction with air

Explanation:

In the measurements of the oscillatory movement in general, the most difficult magnitude of the measurement is the period, by reasoning

* First fight the reaction time of the person, which is approximately 1/10 s, remember that the stopwatch does not stop alone

* Problems to synchronize when starting to measure the oscillation and end point of the oscilloscope, this error needs a good control system to be decreased.

* Effect of friction with air

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A simple machine utilized for completing the works easily. So in order for easy and quick completion of works, these machines may increase the amount of force acting on the object by increasing the velocity or speed of the machine. As velocity term is present so the machines may also change the directions of force acting on the object of concern to do the work soon just like pulley.

Also other way of completing the work with less input force requirement is by increasing the distance or area of action for the force acting on it. As work done is a measure of acting force on a region multiplied with the displacement occurred with that force.

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