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ololo11 [35]
3 years ago
12

Explain what happened to each ball. Why you think it happened? Replay the experiment

Physics
2 answers:
raketka [301]3 years ago
7 0

A dilation can elongate or diminish a figure. It can create an image the same as the original by using a scale factor. The scale factor of a dilation is the ratio of corresponding side lengths.  To dilate a polygon, multiply the coordinates of each vertex by the scale factor k and connect the vertices.

jeka943 years ago
4 0

Answer/Explanation:

A dilation can elongate or diminish a figure. It can create an image the same as the original by using a scale factor. The scale factor of a dilation is the ratio of corresponding side lengths. To dilate a polygon, multiply the coordinates of each vertex by the scale factor k and connect the vertices.

Cheers

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A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid
igor_vitrenko [27]

Answer:

a) attractiva, b) dF = k \frac{Q_1 \ dQ_2}{dx}, c)  F = k Q_1 \frac{Q_2}{d \ (d+L)}, d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

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where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = k \ Q_1 \int\limits^{d+L}_d     {\frac{1}{x^2} } \, dQ_2

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = k \ Q_1 \lambda \int\limits^{d+L}_d  \, \frac{dx}{x^2}

         F = k Q1 λ (-\frac{1}{x})  

we evaluate the integral

        F = k Q₁ λ (- \frac{1}{d+L} + \frac{1}{d} )

        F = k Q₁ λ  ( \frac{L}{d \ (d+L)})

we change the linear density by its value

      λ = Q2 / L

       F = k Q_1 \frac{Q_2}{d \ (d+L)}

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   \frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}

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the sign indicates that the force is attractive

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