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Fofino [41]
3 years ago
8

A particle with charge 5 . 0- µ C is placed at the corner of a cube. (Physics Help)? A particle with charge 5.0-µC is placed at

the corner of a cube. The total electric flux in N * m^2 C through all sides of the cube is: B) 7.1 x 10^4 D) 1.4 x 10^5
Physics
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Gauss law states that the electric flux through any closed surface is proportional to the net electric charge inside the surface. This is expressed mathematically in the form of:

Φ = Q / εo

Where,

Φ = the electric flux = unknown (which we have to find for)

Q = the net electric charge = 5.0 µC = 5 E-6 C

εo = the permittivity of free space = a constant value = 8.85 E-12 C^2 / N m^2

Plugging in the values into the equation will result in:

Φ = 5 E-6 C / (8.85 E-12 C^2 / N m^2)

Φ = 564,971.75 Wb = <span>5.6 x 10^5 Wb </span>

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Answer:

P=1362\ W

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Explanation:

Given:

initial temperature of water, T_i=18^{\circ}C

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desity of water, \rho=1\ kg.L^{-1}

So, the givne mass of water, m=1\ kg

enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

where:

m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

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Now the power rating of the boiler:

P=\frac{Q_s+Q_v}{t}

P=\frac{342760+1128200}{1080}

P=1362\ W

Now the time required to heat to boiling point form initial temperature:

t'=\frac{Q_s}{P}

t'=\frac{342760}{1362}

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