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3 years ago

A 0.900-V potential difference is maintained across a 1.5m length of 2

Physics

1 answer:

Daniel [21]3 years ago

3 0

Answer:

I = 6.42 A

Explanation:

Given that,

Potential difference, V = 0.9 V

Length of the wire, l = 1.5 m

Area of cross section, $A=0.6\ mm^2=6\times 10^{-7}\ m^2$

We need to find the current in the wire. Let I is current. We can find it using Ohm's law as follows :

V = IR

Where R is the resistance of the wire

$I=\dfrac{V}{R}\\\\I=\dfrac{V}{\rho \dfrac{l}{A}}\\\\I=\dfrac{0.9}{5.6\times 10^{-8}\times \dfrac{1.5}{6\times 10^{-7}}}\\\\I=6.42\ A$

So, the current in the wire is 6.42 A.

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Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

3 years ago

If someone said that instruments that produce sounds by vibrating themselves are called aerophones, would that be a true or fals

AfilCa [17]

If someone said that instruments that produce sounds by vibrating themselves are called aerophones. <u>False</u>

<h3>What is aerophones?</h3>

An aerophone is a musical instrument that makes sound primarily by forcing a body of air to vibrate without the use of strings or membranes, and without the vibration of the instrument itself significantly enhancing the sound.

<h3>What is aerophone instrument?</h3>

Any of a group of musical instruments known as aerophones, in which the initial sound is produced by a mass of vibrating air. Instruments that don't fit into any of these categories, such the bull-roarer and the siren, are also included in the basic types, along with woodwind, brass, and free-reed instruments.

<h3>What are African aerophones?</h3>

Aerophones, which create music by vibrating columns of air, are among the instruments used in Africa. Flutes, reed pipes, and trumpets and horns are the three major types of African aerophones.

To know more about aerophone instrument visit:

brainly.com/question/14601234

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4 0

1 year ago

A 0.51-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to it

den301095 [7]

Answer:

487.23 N/m

Explanation:

Given:

mass of metal sphere 'm'= 0.51kg