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Engineering

2 answers:

wel2 years ago

6 0

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german2 years ago

5 0

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A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur

Lina20 [59]

Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and increasing the capacitance proportionally to the dielectric constant of the Teflon.

8 0

3 years ago

Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1

steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

$z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}$

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

$50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$ 50 m + 30.704358 m + 20.4081633 m = 10.234786 m + $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$