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Alinara [238K]
2 years ago
6

The ability to adjust to an artificially displaced or even inverted visual field is called

Physics
1 answer:
egoroff_w [7]2 years ago
3 0

Answer:

méthode if séparation muddy water

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A less common type of radioactive particle emitted is the positron. If potassium-38 forms argon-38 by positron emission, the num
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A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to
Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV  

V = W/q    ....(1)

Using equation (1), the potential difference between points A and B is:

V_{1}=\frac{W_{1} }{q}

Substitute the suitable values in the above equation.

V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

V_{2}=\frac{W_{2} }{q}

Substitute the suitable values in the above equation.

V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

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explain how a mutation in an individual's DNA occurs, and describe the effect it may have on the individual's traits.
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Read 2 more answers
What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

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