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sp2606 [1]
4 years ago
15

Select the options that best complete the statement.Positively charged particle trajectories(always, never, the same as)follow e

lectric field lines, because(electric field lines are defined by the path positive test charges travel.,the particle velocities may or may not be in the same direction as the electric field lines.,positive charges repel each other and, therefore, are repelled by electric field lines.,the electric force on a positively charged particle is in the same direction as the electric field.)
Physics
1 answer:
SpyIntel [72]4 years ago
3 0

Answer:

a) always. b) electric field lines are defined by the path positive test charges travel.

Explanation:

By convention, field lines always follow the direction that it would take a positive test charge  (small enough so it can´t disrupt the field created by a charge distribution), under the influence of an electric field, at the same point where the test charge is located.

So any positive charge, subject to an electric field influence, moves along the field line that passes through its current position, in the same way that a positive test charge would.

We could say also that the electric force on a positively charged particle is in the same direction as the electric field that produces that force (due to some charge distribution) , which is true, but it doesn´t explain why.

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∅ = 89.44°

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In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

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Projection angle ∅ = ?

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where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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