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Karo-lina-s [1.5K]

3 years ago

9

You need to prepare a 0.199-mm-diameter tungsten wire with a resistance of 3.09 kÏ. how long must the wire be? the resistivity o

f tungsten is 5.62 Ã 10-8 ÏÂ·m.

1 answer:

jekas [21]3 years ago

7 0

<span>1.71x10^3 meters. The resistivity of a conductor with an uniform cross section is expressed as R = pl/A where p = resistivity l = length of conductor A = cross sectional area of conductor. Solving for l gives R = pl/A RA = pl RA/p = l The cross section is defined a A = pi r^2, so A = 3.14159 * (9.95x10^-5 m)^2 A = 3.14159 * 9.90025x10^-9 m^2 A = 3.11025x10^-8 m^2 Now substitute the known values into the formula, giving: RA/p = l 3090 ohm * 3.11025x10^-8 m^2 / 5.62x10^-8 ohm * m = l 1.71009x10^3 m Rounding to 3 significant places gives 1.71x10^3 meters.</span>

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When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

igomit [66]

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

The level at which water is been released is $h = 220 \ m$

The efficiency is $\eta =$ 0.90

The electric power output is mathematically represented as

$P = \frac{PE_l - PE _o}{t}$

Where $PE_l$ is the potential energy at level h which is mathematically evaluated as

$PE_l = mgh$

and $PE_o$ is the potential energy at ground level which is mathematically evaluated as

$PE_o = mg(0)$

$PE_o = 0$

So

$P = \frac{mgh}{t}$

here $m = V * \rho$

where V is volume and $\rho$ is density of water whose value is $\rho = 1000 kg/m^3$

So

$P = \frac{(\rho * V) * gh}{t}$

$P = \frac{V}{t} * gh \rho$

substituting values

$P =690 * 9.8 * 220 * 1000$

$P =1.488*10^{9} \ W$

The maximum possible electric power output is

$P_{max} = P * \eta$

substituting values

$P_{max} =1.488*10^{9} * 0.90$

$P_{max} =1.339*10^{9} \ W$

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2 years ago

Which best describes the electric field created by a positive charge?

Vlad [161]

Its rays point away from the charge

3 0

2 years ago

Read 2 more answers

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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2 years ago

A 5 kg ball is sitting on top of a hill. It has a Potential Energy of 6000J. What is the height of the ball?

natali 33 [55]

Explanation:

mass=5 kg

potential energy=6000j

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2 years ago

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Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

7 0

2 years ago