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Snezhnost [94]
4 years ago
6

A common demonstration involves charging a rubber balloon, which is an insulator, by rubbing it on your hair and then touching t

he balloon to a ceiling or wall, which is also an insulator. Because of the electrical attraction between the charged balloon and the neutral wall, the balloon sticks to the wall. Imagine now that we have two infinitely large, flat sheets of insulating material. One is charged, and the other is neutral. If these sheets are brought into contact, does an attractive force exist between them as there was for the balloon and the wall?
Physics
1 answer:
baherus [9]4 years ago
5 0

Answer:

here there is also an electric attraction force

Explanation:

In the insulating bodies the charges are not mobile, so when one of the body is charged periodically or gained some electrical charges.

When the body is neutral the net number of electric charges is zero. When the loaded body touches it, it repels the charges of the same sign whereby the induced load is of a different sign and the two adhered sheets are attracted.

With this explanation it is equivalent to that of the ball and the wall, so here there is also an electric attraction force

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Definition of measurements?​
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Tarzan, who weighs 825 N, swings from a cliff at the end of a 19.7 m vine that hangs from a high tree limb and initially makes a
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Answer:

a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,

c)  F = 281.70 N, d)    θ = 338.58º , e)  a = 3,588 m / s² , f)  θ = 201.45º

Explanation:

For this exercise we will use Newton's second law on each axis

X axis

         -Tₓ = m aₓ

Y Axisy

          T_{y} –W = m a_{y}

Let's use trigonometry to find the components of force

          sin 21.5 = Tₓ / T

          cos 21.5 = T_{y} / T

          Tₓ = T sin 21.5

          T_{y} = T cos 21.5

          Tₓ = 768 sin 21.5 = 281.47 N

          T_{y} = 768 cos 21.5 = 714.56 N

a) the force of the rope on Tarzan is

          T = (281.47 i ^ + 714.56 j ^) N

b) The net force is the subtraction of the tension minus the weight of Tarzan

Y  Axis   F_net = 714.56 - 825 = -110.44 N

              F_net = (281.47 i ^ - 110.44 j ^) N

c) Let's use Pythagoras' theorem

      F = √ (Fₓ² + T_{y}²)

      F = √ (281.47² + 110.44²)

      F = 281.70 N

d) Let's use trigonometry

     tan θ = F_{y} / Fₓ

      θ = tan⁻¹ F_{y} / Fₓ

      θ = tan⁻¹ (-110.44 / 281.47)

       θ = -21.42º

This angle is average clockwise, for counterclockwise measurement

       θ = 360 - 21.42

       θ = 338.58º

Acceleration

X axis

             Tₓ = m aₓ

             aₓ = Tₓ / m

The mass of Tarzan is

             m = W / g

             m = 825 / 9.8 = 84.18 kg

             

             aₓ = 281.47 / 84.18

             aₓ = -3.34 m / s2

Y Axis

            T_{y}-W = m a_{y}

            a_{y} = (T_{y} -W) / m

            a_{y} = (714.56-825) / 84.18

            a_{y} = - 1,312 m / s²

Acceleration Module

             a = √ aₓ² + a_{y}²

             a = √ (3.34² +1.312²)

             a = 3,588 m / s²

The angle

          θ = tan⁻¹ a_{y} / aₓ

          θ = tan⁻¹ (-1312 / -3.34)

          θ = 21.45º

Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis

          θ = 180 + 21.45

          θ = 201.45º

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