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babunello [35]
2 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]2 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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Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

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2 years ago
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The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

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Which statements can be inferred from the Paleozoic era time scale? There was volcanic activity during the Paleozoic era. Dinosa
Alchen [17]

Answer:

A,C,E

Explanation:

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1 year ago
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The descriptions below explain two ways that water is used by plants on a sunny day. I. In a process called transpiration, some
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In photosynthesis, the water is being used to create food for the plant (Glucose). In transpiration the water is going from a liquid to a gas that's being released.
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A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0
2 years ago
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