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777dan777 [17]
3 years ago
11

A student is performing experiments on a particular substance. Which

Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer: A.

Explanation:

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a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
3 years ago
A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all
timurjin [86]

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, a=5\times 10^{-4}\ m^2

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

P=\dfrac{F}{A}

P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}

P=1.4\times 10^5\ Pa

We know that the atmospheric pressure is given by :

P_o=1.013\times 10^5\ Pa

So, the pressure is given by :

p=P+P_o

p=1.4\times 10^5+1.013\times 10^5

p=2.41\times 10^5\ Pa

Hence, this is the required solution.

7 0
3 years ago
When an experiment is replicated, how should the results of the two experiments compare?
olasank [31]
When the experiment is replicated, this means the conduction of the second experiment, should be related, or similar, to your first results. 
4 0
3 years ago
Read 2 more answers
A group of Advanced Open Water Divers plans to make two dives. The first dive is on a reef in 90 feet of water for 20 minutes. T
Kaylis [27]

Answer:

The ending pressure is R.

Explanation:

This is Recreational scuba diving (R). According to standards, this diving has prescribed depth limit of 130 ft. This means that the limit must include a depth that is not greater than 130 ft while using only compressed air and not even requiring a decompression stop.

Now, from the question, we can see that that there was a prescribed limit of 60ft of water with a planned bottom time of 30 minutes.

Thus, the ending pressure is R.

5 0
3 years ago
The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much
aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}

Upon substituting our given values, we will get:

\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}

\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365

Taking square root of both sides, we will get:

\frac{T_1}{T_2}=\sqrt{0.0073878361246365}

\frac{T_1}{T_2}=0.0859525225030452495

\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}

\frac{T_2}{T_1}=11.634329870476699\approx 11.634

T_2=11.634\cdot T_1

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0
3 years ago
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