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Mekhanik [1.2K]
1 year ago
10

A cubical surface surrounds a point charge q . Describe what happens to the total flux through the surface if (c) the surface is

changed to a sphere
Physics
1 answer:
valentina_108 [34]1 year ago
3 0

Answer:

Gauss law states that the electric flux is defined as the electric field multiplied by the area of the surface in a plane perpendicular to the field.

Explanation:

Mathematically,

Φ=Q ϵo

Where;

Q is enclosed charge

ϵo is the permittivity of the free space

According to Gauss law, which states that the electric flux is defined as the electric field multiplied by the area of the surface in a plane perpendicular to the field.

Φ=Q ϵo

Where;

Q is enclosed charge

ϵo is the permittivity of the free space

If the cube is transformed into a sphere the total flux in the electric field remains unchanged or remains the same. This is because the gaussian law does not postulate that electric flux is dependent on the object in a plane. Hence, the transformation of the cube to a sphere does not affect the electric flux generated in the field.

To learn more about how the total flux through a sphere relates to the surface change, click brainly.com/question/4362789

#SPJ4

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The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
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Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

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