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3 years ago

Choose all statements that are true about a GFCI outlet.

Physics

1 answer:

saveliy_v [14]3 years ago

5 0

GFCI outlets are found in wet areas is the true statement about a GFCI oulet

Answer: Option D.

Explanation:

GFCI stands for Ground Fault Circuit Interrupter and also named as RCD (Residual Current Device). This is a kind of circuit breaker that cuts off electricity after detecting an imbalance between output and input currents. The switch protects household lines and sockets against overheating and possible fire.

GFCI protects the people and is mostly used in bathrooms or kitchen where electrical equipment is used. Also, it is available where the human body can get into contact with the floor or metal fixings that provide an alternative route to power in the event of a fault. These outlets are available in two versions: the circuit breaker installed in the control panel and receptacle type installed into the electrical box.

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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

Tpy6a [65]

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

7 0

3 years ago

Explain the theory of Plate Tectonics. b) How does it account for the Mid-Atlantic Ridge? c) How does it account for volcanic is

hjlf

Answer:

A) The continents and ocean basins undergo continuous change. Both are parts of lithospheric plates that move against each other. B) Divergent plate in Mid-Atlantic Ridge with material flowing into the ocean. C) A plate moved over a stationary site of magma upwelling "Hot Spot" and created a volcanic island chain over the time

Explanation:

A) The basic thought is, that instead of being permanent fixtures of the earth's surface, the continents and ocean basins undergo continuous change. Both are parts of lithospheric plates that move against each other, and in the process new crust is created at midoceanic ridges (spreading centers), and old crust is consumed at convergent plate boundaries (subduction zones).

B) There are basically three different types of plate boundaries:

Divergent boundaries -- where new crust is generated as the plates pull away from each other.

Convergent boundaries -- where crust is destroyed as one plate dives under another.

Transform boundaries -- where crust is neither produced nor destroyed as the plates slide horizontally past each other.

The best known of the divergent boundaries is the Mid-Atlantic Ridge. This submerged mountain range, which extends from the Arctic Ocean to beyond the southern tip of Africa, is but one segment of the global mid-ocean ridge system that encircles the Earth.

C) The linear arrangement of many seamounts indicates that they formed because the plate moved over a stationary site of magma upwelling, a so called mantle "Hot Spot". Seamounts are submarine volcanoes that may finally build above the water level, in which case they are called islands. If seamounts rise above sea level (due to buildup of material in a cone or upwelling mantle pushes up plate), they are subject to wave erosion and colonization by reefs, with both processes tending to create a flat top on the original volcanic cone.

5 0

3 years ago

Help ASAP

SIZIF [17.4K]

I think all of those are examples

4 0

3 years ago

What types of elements are useful for dating materials?

Semenov [28]

Well im not sure if this is the correct dating materials but here are some examples of Fundamentals of radiometric datingRadioactive decay.
Accuracy of radiometric dating.
Closure temperature.
The age equation.
Uranium–lead dating method.
Samarium–neodymium dating method.
Potassium–argon dating method.
Rubidium–strontium dating method.

3 0

3 years ago

Read 2 more answers