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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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Stolb23 [73]

Answer:

m = 0.111 kg

Explanation:

Heat required to release from the body of the person when his temperature cool down by 1 degree C is given as

Q = m s\Delta T

here we know that

m = 70 kg

s = 3840 J/kg K

\Delta T = 1.00^o C

now we know that

Q = (70 kg)(3840 J/kg ^oC)(1 ^o C)

Q = 268800 J

now the same heat is used to vaporize water of the body

so it is given as

Q = mL

268800 = m(2.42 \times 10^6)

m = 0.111 kg

7 0
2 years ago
A steel cable lifting a heavy box stretches by ΔL . In order for the cable to stretch by only half of ΔL , by about what factor
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Answer:

2.0

Explanation:

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6 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
2 years ago
For metalloids on the periodic table, how do the group number and the period number relate?
lapo4ka [179]
Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right
3 0
2 years ago
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