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Naya [18.7K]
3 years ago
8

A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the elect

ric field established between the plates
Physics
1 answer:
FromTheMoon [43]3 years ago
7 0

Complete question:

A uniform electric field is created by two parallel plates separated by a

distance of 0.04 m. What is the magnitude of the electric field established

between the plates if the potential of the first plate is +40V and the second

one is -40V?

Answer:

The magnitude of the electric field established between the plates is 2,000 V/m

Explanation:

Given;

distance between two parallel plates, d = 0.04 m

potential between first and second plate, = +40V and -40V respectively

The magnitude of the electric field established between the plates is calculated as;

E = ΔV / d

where;

ΔV is change in potential between two parallel plates;

d is the distance between the plates

ΔV = V₁ -V₂

ΔV = 40 - (-40)

ΔV = 40 + 40

ΔV = 80 V

E = ΔV / d

E = 80 / 0.04

E = 2,000 V/m

Therefore, the magnitude of the electric field established between the plates is 2,000 V/m

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The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

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Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

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3 years ago
You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T
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Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

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     d = 1.67 10⁻³ mm

    d = 1-67 10⁻³ m

Let's calculate the angle

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First order

    θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

    θ₁ = sin⁻¹ (3.93 10⁻¹)

    θ₁ = 23.14 °

Second order

     θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

     θ₂ = sin⁻¹ (0.786)

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Both nuclear fission and fusion processes involve nuclei of atoms.

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Frozen water that is ice has less density than liquid water. The density of ice is 0.9167 g/cm^3 and the density of the water is 0.9998 g/cm^3.

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