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Genrish500 [490]

3 years ago

14

When a voltage difference is applied to a piece of metal wire, a 10.0 mA current flows through it. If this metal wire is now rep

laced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire?

1 answer:

Mrrafil [7]3 years ago

6 0

Answer:

10 mA

Explanation:

We have given when a potential difference is applied then a current of 10 mA is flows so current i = 10 mA

According to ohms law voltage V =i R

And resistance is given as $R= \frac{\rho l}{A}$

From the relation it is clear that resistance is inversely proportional to the area

As in new case the wire have twice diameter means 4 times the area so the resistance will decrease by 4 times from the previous resistance

As current $i=\frac{V}{R}$ current is inversely proportional to resistance so current will become 4 times of previous value

So current = 10×4=40 mA

You might be interested in

Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar ( $a_{D}$ ), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities ( $v_{D}$ , $\omega_{BD}$ ), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations ( $a_{D}$ , $\alpha_{BD}$ ), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$ (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$ (2)

<h3>Accelerations</h3>

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$ (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$ (4)

If we know that $\theta = 60^{\circ}$ , $\gamma = 19.889^{\circ}$ , $r_{BD} = 10\,in$ , $\omega_{AB} = 16\,\frac{rad}{s}$ , $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$ , then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

$v_{D}+3.402\cdot \omega_{BD} = -41.569$ (1)

$9.404\cdot \omega_{BD} = -24$ (2)

$v_{D} = -32.887\,\frac{in}{s}$ , $\omega_{BD} = -2.552\,\frac{rad}{s}$

<h3>Accelerations</h3>

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$ (3)

$-9.404\cdot \alpha_{BD} = -687.264$ (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$ , $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0

1 year ago

These two pls :)))) ill mark brainliest :)

Anna71 [15]

Answer:

Bowling Ball: weight on Earth = 49 N

Textbook: Mass = 2 kg; weight on the moon = 3.2 N

Large dog: weight on Earth = 490 N; weight on the moon = 80 N

Law of Universal Gravitation: $F_{G}=\frac{Gm_{1}m_{2}}{r^{2}}$

$F_{G}$ = gravitational force (Newtons/N)

<em>G</em> = gravitational constant, 6.67430 × 10¹¹ $\frac{N*m^{2}}{kg^{2}}$

<em>m</em>₁ and <em>m</em>₂ = masses of two objects (kilograms/kg)

<em>r</em>² = square of distance between centers of the two objects (meters/m)

Have a fantastic day!

4 0

2 years ago

Considering how the parts of a system work together and affect one another, other systems, and the environment is called _______

-BARSIC- [3]

Answer:

Hi

before I answer a question I think very deeply and try my best, hope it helps...

As you know there are many different types of systems. For example, The solar system, galaxies, quantum systems, atoms, molecules, orchestras, nervous system, etc, things you may not have even considered a system. To get to the basis of a system we must first understand what a system is then we will show some examples. A system is a group of Parts (parts could mean anything even dark energy and dark matter) that work together to accomplish something. For example, your body has many many trillions of cells that all try to accomplish the functions of humans which include thinking, moving, breathing, circulation, etc. Cells in turn are a system that have counterparts called organelles that accomplish harvesting energy, making new proteins, getting rid of waste, and so on. These are some systems which we highly dependent upon.

Well i hope it helped

Spiky Bob your answerer

8 0

2 years ago

If a book has a mass of 3 kg, what is the book's weight in N?

sesenic [268]

Answer:

29.4 N

Explanation:

F = ma

F = (3 kg) (9.8 m/s²)

F = 29.4 N

5 0

3 years ago

20. Consider a model steel bridge that is 1/100 the exact scale of the real bridge that is to be built. a. If the model bridge w

Veseljchak [2.6K]

The model bridge captures all the structural attributes of the real bridge, at a reduced scale.

Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.

Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.

Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.

6 0

3 years ago