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2 years ago

Find the electric flux through a cylindrical Guassian surface produced by a point charge at base.

Physics

1 answer:

Simora [160]2 years ago

7 0

Answer:

Electric flux will be $\Phi =\frac{q}{\epsilon _0}$

Explanation:

Let the point charge enclosed by Gaussian surface is q

We have to find the electric flux through this Gaussian surface

According to Gauss law electric flux through any closed surface is equal to $\frac{1}{\epsilon _0}$ times of charged enclosed by the surface.

So electric flux will be $\Phi =\frac{q}{\epsilon _0}$

Gauss law is only valid for closed surface it is not applied for open surface

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Answer:

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Explanation:

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Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến

Mila [183]

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

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Now, we will find the energy required to heat the water between given temperature limits:

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where,

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C = specific heat capacity of water = 4182 J/kg.°C

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Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

$Power = P = \frac{E}{t}$

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

$P = \frac{2.927\ x\ 10^6\ J}{1200\ s}$

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2 years ago

Why is centrifugal force is a fake force?<br>

mash [69]

Answer:

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2 years ago

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β = 0,5/0,7

tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰

cos β = 0,81

d = √ (0,5)² + (0,7)² d1stance between charges

d = √0,25 + 0,49

d = √0,74 m

d = 0,86 m

Now Foce between two charges is:

F = K* q₁*q₂/ d² (1)

Where K = 9*10⁹ N*m²/C²

q₁ = 2,5* 10⁻⁹C

q₂ = 3,0*10⁻⁹C

d² = 0,74 m²

Plugging these values in (1)

F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹ [N]

And Fx = F*cos β

Fx = 91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹ [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹ [N]

F = 147,78*10⁻⁹ [N]

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3 years ago