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vovikov84 [41]
2 years ago
6

The electric field in a particular space is = (x + 3.6) N/C with x in meters. Consider a cylindrical Gaussian surface of radius

16 cm that is coaxial with the x axis. One end of the cylinder is at x = 0. (a) What is the magnitude of the electric flux through the other end of the cylinder at x = 3.7 m? (b) What net charge is enclosed within the cylinder?
Physics
1 answer:
Dvinal [7]2 years ago
6 0

Answer:

a) Ф = 0.016 N / C m , b) q_{int} = 0.14 10⁻¹² C

Explanation:

a) For this problem we use Gauss's law

          Ф = E .ds = q_{int} /ε₀

The camp is in the x direction so it has no flow through the cylinder walls.

          Ф = E A

         

The area of ​​a circle is

           A = π r

          Ф = E π r

          Ф = (x- 3.6) r

Let's calculate

          Ф = (3.7 -3.6) 0.16

          Ф = 0.016 N / C m

     

b) we clear from Gauss's law

             q_{int} = Ф ε₀

 

Where the flow is on both sides, on the face at x = 0 the flow is zero

             q_{int} = 0.016 8.85 10⁻¹²

             q_{int} = 0.14 10⁻¹² C

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Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
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ΔTf = Kf×m
ΔTf = depression in the freezing point.
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= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
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7 0
2 years ago
Read 2 more answers
You are using a lightweight rope to pull a sled along level ground. The sled weighs 485 N, the coefficient of kinetic friction b
Bogdan [553]

Answer:

N=459.01N

Explanation:

According to Newton's first law:

N+F_y-W=0

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Replacing and solving for N:

N=W-Fsin12^\circ\\N=485N-(125N)sin12^\circ\\N=459.01N

5 0
2 years ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
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Answer:

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\hat B = \hat v \times \hat r

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\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

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3 0
2 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

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Required

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1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

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R2 = 129765/428

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