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alina1380 [7]
2 years ago
9

A cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces.

No other charges are nearby. (ii) Through how many of the cube's faces is the electric flux zero? Choose from the same possibilities as in part (i).
Physics
1 answer:
laiz [17]2 years ago
7 0

Answer:

The electric flux is zero through four cube surfaces given that a cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces.

Explanation:

Assuming the charged filament is quite long and you are not near the edges, the two opposing sides that the filament travels through have no flux. If the charge filament is long, which you may assume is indefinitely long, then there is the equal amount of charge on the left and right of where you are, therefore the electric field has no preference for left or right. This implies that the electric field can only travel in or out of the filament. No field lines run through the two faces of the cube that the filament goes through if the electric field is not moving left or right. There are electric field lines on the four sides of the filament.

To learn more about cubical gaussian surface and electric flux. Click brainly.com/question/13003911

#SPJ4

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Explain what happens to the energy in a group in a system if one object loses energy according to the Law of Conservation of Ene
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energy never disappears, for example, if you give some kinetic energy to a ball and it stops few seconds later, friction steals this energy to ground which ball was going on. "Law of Conservation of Energy" tell us that energy can't disappear

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Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration
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it's acceleration is 0

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since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0
3 years ago
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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4 0
2 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
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