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lidiya [134]
3 years ago
12

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

um plates, each 21 cm in diameter, separated by 1.0 cm . Part A How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3× 10 6 N/C of the field causes a spark. Express your answer with the appropriate units. q max q m a x = nothing nothing SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
Alex777 [14]3 years ago
7 0

Answer:

0.92 μC

Explanation:

In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:

E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C

or 0.92 μC

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I’ll give brainliest, help!!
antiseptic1488 [7]

Answer:

A,W = 40 J

Explanation:

A, W =FS

W=10N(4m)

W = 40 J

8 0
2 years ago
When is the universal theme of a story often revealed?
elena-14-01-66 [18.8K]
It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>
7 0
3 years ago
Read 2 more answers
Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation s
77julia77 [94]

Answer:

a) dB / dA = 2 ,

b) Network B     Network A

        2                         1

        4                         2

        6                         3

Explanation:

a) The expression for grating diffraction is

         d sin θ = m λ

where d the distance between two slits, λ the wavelength and m an integer that represents the diffraction range

In this exercise we are told that the two spectra are in the same position, let's write the expression for each network

Network A

m = 1

         sin θ = 1 λ / dA

Network B

m = 2

        sin θ = 2 λ / dB

they ask us for the relationship between the distances, we match the equations

            λ / dA) = 2 λ / dB

            dB / dA = 2

         

b) let's write the equation of the networks

         sin θ = m_A  λ / dA

         sin θ = m_B  λ / dB

we equalize

           m_A  λ/ dA = m_B  λ / dB

we use that

          dB / dA = 2

           m_A 2 = m_B

therefore the overlapping orders are

Network B     Network A

   2                         1

   4                         2

    6                       3

4 0
3 years ago
A 0.03-kg bullet is fired with a horizontal velocity of 470 m/s and becomes embedded in block B which has a mass of 3 kg. After
Gala2k [10]

Answer with Explanation:

We are given that

Mass of bullet,m_1=0.03 kg

u_1=470 m/s

m_2=3 kg

\mu_k=0.2

m_3=30 kg

We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.

According to law of conservation of momentum

m_1u_1=(m_1+m_2)v

0.03(470)=(0.03+3)v

v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s

Hence, the velocity of the bullet and block B after the first impact=4.65 m/s

According to law of conservation of momentum

(m_1+m_2)v=(m_1+m_2+m_3)V

(0.03+3)\times 4.65=(0.03+3+30)V

V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}

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3 0
4 years ago
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The decomposition of ammonia to the elements is a first-order reaction with a half-life of 200 s at a certain temperature. How l
Feliz [49]

Based on the half-life of the reaction, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

<h3>What is the half-life of a substance?</h3>

The half-life of a substance is the time taken for half the amount of atoms present in that substance to decay.

The half-life of the reaction is 200 seconds.

After, one half-life, pressure reduces to 0.0500 atm

After, two half-lives, pressure reduces to 0.0250 atm

After, three half-lives, pressure reduces to 0.01250 atm

After, four half-lives, pressure reduces to 0.00625 atm

Time taken = 4 * 200 = 800 seconds

Therefore, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

Learn more about half-life at: brainly.com/question/26689704

#SPJ1

7 0
2 years ago
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