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3 years ago

What is the fault-current circuit breaker? Describe its function.

Physics

2 answers:

Nimfa-mama [501]3 years ago

4 0

Answer:

A circuit breaker is a switching mechanism that interrupts the current that is irregular or fault. It is a mechanical system that interferes with high magnitude (fault) current flow and also performs a transfer operation. Hope this helps :)

Explanation:

AleksandrR [38]3 years ago

4 0

Answer:

A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by an overcurrent or short circuit.

Explanation:

Its basic function is to interrupt current flow after protective relays detect a fault.

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Two particles with oppositely signed charges nC are placed at two of the vertices of an equilateral triangle with side length 3

babymother [125]

The magnitude of the electric field at the third vertex of the triangle is determined as zero.

<h3>Electric field at the third vertex of the triangle </h3>

The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;

E = E(13) + E(23)

E = (kq₁)/r² + (kq₂)/r²

where;

q1 is positive charge
q2 is negative charge

E = (kq₁)/r² - (kq₂)/r²

E = 0

Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

5 0

2 years ago

As the temperature increases, materials with

Andrews [41]

Answer:large

Explanation:

As the temperature increases, materials with large coefficients of linear expansion increases a lot in size

7 0

3 years ago

1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's

sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth = 5.97 x 10²⁴kg

Distance = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

F = $\frac{Gm_{1} m_{2} }{r^{2} }$

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is the distance

F = $\frac{6.67 x 10^{-11} x 7.34 x 10^{22} x 5.97 x 10^{24} }{(3.8 x 10^{5})^{2} }$

F = $\frac{2.92 x 10^{35} }{1.44 x 10^{11} }$ = 2.03 x 10²⁴N

8 0

3 years ago

it took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 miles per hour. How many miles lie

Art [367]

Given:

Time: 3.5 hrs

Velocity: 120 miles/hr

Now Distance= Speed × Time

Now Velocity and speed have the same magnitude. Velocity being a vector quantity has a definite direction. Whereas speed is a scalar quantity,it indicates only the magnitude an doesn't define any direction.

Hence Distance = Velocity x time

Distance = 3.5 × 120 = 420 miles

7 0

3 years ago

Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+

babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• $\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j$ with $0\le x\le2$ and $0\le y\le6-x$ ;

• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

• $\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k$ with $0\le u\le\frac\pi2$ and $0\le y\le6-2\cos u$ .

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

$\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k$

$\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j$

$\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i$

$\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j$

$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0$

$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

$\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8$

$\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz$

$=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0$

$\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du$

$=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi$

$\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du$

$=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24$

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV$

where <em>R</em> is the interior of <em>S</em>. We have

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7$

The integral is easily computed in cylindrical coordinates:

$\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2$

$\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3$

as expected.

4 0

3 years ago