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babunello [35]
3 years ago
5

You are working on charge-storage devices for a research center. Your goal is to store as much charge on a given device as possi

ble. The facilities allow you to generate almost any potential difference V you need, but you are restricted to using a single parallel-plate capacitor. The area and separation distance of the plates are fixed, and the dielectric materials available to you are paper (κ = 3.0, Emax=4.0×107V/m), Mylar (κ = 3.3, Emax=4.3×108V/m), quartz (κ = 4.3, Emax=8×106V/m), and mica (κ = 5, Emax=2×108V/m).
Part A. What properties of these materials must you consider in choosing the best dielectric for your needs? Check all that apply.
a. mass density
b. dielectric constant breakdown threshold
c. electric conductivity
Part B. Rank the materials in order of their ability to meet your needs, first choice first. Rank the materials from the most appropriate to the least appropriate. To rank items as equivalent, overlap them. HelpReset paperquartzmicaMylar Least appropriateMost appropriate The correct ranking cannot be determined.
Physics
1 answer:
zheka24 [161]3 years ago
8 0

Answer:

Part A the answer is the dielectric constant.

Part B  Mica- mylar- paper- quartz

Explanation:

The capacity of a capacitor is given by

           C = ε ε₀ A / d

Where the dielectric constant (ε) is the value of the material between the plates of the capacitor, we see that as if value increases the capacity also increases.

Another magnitude that we must take into account that the maximum working voltage, the greater the safer is the capacitor

the flexibility of the material must also be taken into account

Part A the answer is the dielectric constant.

Pate B order the materials from best to worst

Mica. The best ever

Mylar Flexible

Paper Low capacity, low working voltage, flexible

Quartz high dielectric, but brittle

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What is the difference between chemical change and a physical change?
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3 0
3 years ago
Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0
lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

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2 years ago
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Oksanka [162]

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

Q_1 = Q_2

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

Q = VA

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

Q_2 = 1.2*10^{-4}m^3/s

d = 0.021m

r = \frac{0.021}{2} = 0.0105m

Since there is continuity we have now that,

V_1A_1 = Q_2

V_1A_1 = 1.2*10^{-4}

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7 0
3 years ago
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

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So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

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

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

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d = 0.144 mm

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So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0
3 years ago
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