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Bad White [126]
3 years ago
12

A point charge q +4 μC is at the center of a sphere of radius 0.4 m. (a) Find the surface area of the sphere. 2.01 (b) Find the

magnitude of the electric field at points on the surface of the sphere. m2 X N/C (c) What is the flux of the electric field due to the point charge through the surface of the sphere? N -m2/c d) Would your answer to part (c) change If the polnt charge were moved so that It was Inslde the sphere but not at Its center? yes no not enough information to decide (e) What Is the net flux through a cube of side 2 m that encloses the sphere?
Physics
1 answer:
vlabodo [156]3 years ago
3 0

Answer:

area  A = 2.01 m² ,  electric fields E = 2.25 10⁵ N/C and flux   Φ = 4.52 10⁵ N m²/C

Explanation:

The electric field of a point charge can be calculated with the equation

       E = k q / r²

Where q is the point charge, r is the distance of the charge to the sphere and k the constant Coulomb is  8.99 10⁹ N m² / C²

a) The surface or area of ​​a sphere is

       A = 4π R²

       A = 4 π 0.4²

       A = 2.01 m²

b) We use the electric field equation

      E = 8.99 10⁹ 4 10⁻⁶ / 0.4²

      E = 2.25 10⁵ N / C

c) The electric flow can be calculated by the equation

     Φ = ∫ E. dA

The point is the scalar product, the electric field lines leave the charge and as the charge is in the center of the sphere, it co-exists with the radii of this, which are the lines of the area, so they are parallel and the scalar product it is reduced to the ordinary product

      Φ = E A

      Φ = 2.25 10⁵ 2.01

      Φ = 4.52 10⁵ N m²/C

d) If we write Gauss's law

       Φ = E A = Qint / ε₀

When analyzing this law the flow is proportional to the value of the charge inside the spheres, regardless of the point where this face is located, so if the charge moves a little it will not change the total flow on the sphere.

e) Let's use Gauss's law for this part

         Φ = Qint / eo

Regardless of the shape of the surface

        Φ = 4 10⁻⁶ / 8.85 10⁻¹²  

        Φ = 4.52 10⁵ N m² / C

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Answer:

B)The motion of water in an ocean current

Explanation:

With respect to measurements, a vector has both a magnitude and a direction. The first three examples (maximum height of a hill, air temperature, and rain accumulation) are magnitudes only. The fourth example (motion of water in an ocean current) is a vector, because it has a magnitude (speed) and a direction (with the current).

4 0
3 years ago
A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s
balandron [24]

Answer:233.23 m/s^2

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

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1.5=0+\frac{gt^2}{2}

t=0.55 s

Initial horizontal velocity at the time of break is given by u

R=u\times t

10=u\times 0.55

u=18.07 m/s

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6 0
3 years ago
Which term below describes a measurement of how hard an object pushes against a surface?
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A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b
Andru [333]

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³       ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

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We know that work done by gas given as

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W= - 0.72 KJ

W= - 720 J    ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390  J

ΔU= 330 J

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3 years ago
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KiRa [710]
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