Which of the following is the method by which air is cooled in the food storage component of a refrigerator?

Two identical air-filled parallel-plate capacitors C1 and C2 are connected in series to a battery that has voltage V. The charge

Llana [10]

Complete question:

Two identical air-filled parallel-plate capacitors C₁ and C₂ are connected in series to a battery that has voltage V. The charge on each capacitor is Qo. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K > 1 is inserted between the plates of capacitor C1, completely filling the space between them. What is the charge on capacitor C1 after the dielectric is inserted? Express your answer in terms of K and Qo.

Answer:

The charge on capacitor after the dielectric is inserted is Qo( k + 1)

Explanation:

Given;

two identical capacitor, C₁ and C₂

charge on each capacitor, Q = Qo

Voltage of the battery, = V

dielectric constant K > 1

After inserting dielectric constant K, the equivalent capacitance in parallel becomes;

C.equ. = KC₁ + C₂

Charge, on capacitor Qo = C.equ.V

Q = V(KC₁ + C₂)

Q = KC₁V + C₂V

since, the charge on each capacitor is Qo, then

Q = KQo + Qo

Q = Qo( k + 1)

Therefore, the charge on capacitor after the dielectric is inserted is Qo( k + 1), in terms of K and Qo.

5 0

3 years ago

Light has wavelength 600 nm in a vacuum. it passes into glass, which has an index of refraction of 1.5. what is the frequency of

Lelu [443]

v = f lambda

in vac ... 3X10^8 = 600x10^-9xf

in glass speed slower, poss 2/3 that of vacuum

7 0

3 years ago

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

4 0

4 years ago

5. Lady Gaga needed someone to buy more meat for her meat dress. The famous

krok68 [10]

Answer:

she can travel 24000 meters she cant reach the store she can get as much as 125 miles

Explanation:

7 0

3 years ago

Examine the graph. Select the statement that best describes the energy change in the particles of a substance during melting.

serg [7]

Answer:

B) Absorbed energy results in the change in potential energy.

Explanation:

i got it wrong on usatestpreo

3 0

3 years ago

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