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neonofarm [45]
3 years ago
6

While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega

tively charged rod is removed. What would happen to the vane?
Physics
1 answer:
Paraphin [41]3 years ago
4 0

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

 

   Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

   In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

   If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

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Verdich [7]
So you can use the equation force = mass x acceleration to do 2 x 5 to get 10 N
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3 years ago
A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the
olasank [31]

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,     \theta = 30^{o}

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin \theta = \frac{1}{2}kx^{2}

       h = \frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}

          = \frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0
3 years ago
A pipe of length 10.0 m increases in length by 1.5 cm when its temperature is increased by 90°F. What is its coefficient of line
azamat

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

  • Original diameter (L₁) = 10 m
  • Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
  • Change in temperature (∆T) = 90 °F
  • Coefficient of linear expansion (α) =?

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

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3 0
1 year ago
What do you think we call this graphical representation based on your prior experience with electric fields and electric field l
slavikrds [6]

Answer:

Explanation:

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Rules for drawing electric field lines

1. Electric field lines are always drawn from High potential to

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2. Two electric field lines can never intersect each other.

3. The net electric field inside a Conductor is Zero.

4. Electric field line from a positive charge is drawn radially outwards and from a negative charge radially inwards.

5. The density of electric field lines tells the strength of the electric field at that region.

6. Electric field lines terminate Perpendicularly to the surface of a conductor.

A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.

So, electric fields are vector quantity due to the fact any student can tell you that a compass is used to determine which direction is north.

Since the compass always point northward, then it has a direction and magnitude and so it is a vector quantity

6 0
3 years ago
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3 years ago
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