1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
aalyn [17]
2 years ago
6

If a dielectric material, such as teflon®, is placed between the plates of a parallel-plate capacitor without altering the struc

ture of the capacitor, how is the capacitance affected?.
Engineering
1 answer:
erastovalidia [21]2 years ago
4 0

The capacitance is affected is that the capacitance increases because of the insertion of Teflon.

<h3>What is capacitance?</h3>

A `is known to be a form of a device that is said to be used to save some amount of charges in any electrical circuit.

Note that the capacitor functions on the principle that the capacitance of a conductor is said to increases if an earthed conductor is  taken close to it.

Hence, The capacitance is affected is that the capacitance increases because of the insertion of Teflon.

See full question below

A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structure of the capacitor. The charge on the capacitor is held fixed. How is the electric field between the plates of the capacitor affected?

The electric field is not altered, because the structure remains unchanged.

The electric field becomes zero after the insertion of the Teflon®.

The electric field decreases because of the insertion of the Teflon®.

The electric field becomes infinite because of the insertion of the Teflon®.

The electric field increases because of the insertion of the Teflon®.

Learn more about  electric field  from

brainly.com/question/14781269

#SPJ1

You might be interested in
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
Which of the following best reflects a shield system?
Dafna11 [192]
C, because a narrow structure evacuation below surface ground isn’t the best and a structure holding forces and isn’t to do with the question at all and d doesn’t matter if there include away or not
8 0
3 years ago
Coefficients of friction always involve how many materials? a. 0 b. 1 c. 2 d. 3
True [87]

Answer:

Answer is c. 2

Explanation:

Since friction is present at the contact  between 2 materials thus coefficient of friction always involve 2 materials

5 0
3 years ago
A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after
Talja [164]
<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

v_i = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = \frac{v_f \:-\:v_i}{t}

where: a - acceleration

v_i - initial velocity

v_f - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a = \frac{1700\:N}{1250\:N}

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>
  • Convert 30 km/hr to m/s

= \frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}

= 8.33 m/s

  • Substitute the value of a, v_i and t

a = \frac{v_f \:-\:v_i}{t}

1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}

(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s

v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s

v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s

v_f\: =\:  21.93\: m/s

  • Convert to km/hr

= \frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}

= 78.95\: km/hr

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
6 0
3 years ago
Other questions:
  • Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the
    6·1 answer
  • "Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."
    15·1 answer
  • What are the four basic parts of process plan
    11·1 answer
  • Create a project named CarDealer that contains a Form for an automobile dealer. Include options for at least three car models. A
    14·1 answer
  • Consider the cascade of the three LTI systems having impulse responses: h-1(t) = e^-tu(t + 3) h_2(t) = rect((1 -1)/2) h_3(t) = d
    8·1 answer
  • When would working with machinery be a common type of caught-in and caught-between<br> hazard?
    6·1 answer
  • How much thermal energy is needed to raise the temperature of 15kg gold from 45⁰ C up to 80⁰ C​
    10·1 answer
  • Which one of these is not a successful budgeting strategy
    5·2 answers
  • Limited time only for christmas give yourself free 100 points Thats what im talking about
    5·2 answers
  • 7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!