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tatuchka [14]
3 years ago
12

Which of the following statements about electric field lines associated with electric charges is false? Electric field lines can

be either straight or curved. Electric field lines can form closed loops. Electric field lines begin on positive chares and end on negative charges. Electric field lines can never intersect with one another.
Physics
1 answer:
Tcecarenko [31]3 years ago
8 0

Answer:

The false statement is 'Electric field lines form closed loops'.

Explanation:

  • Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
  • These lines when close together represents high intensity and when far apart shows low intensity of the field.
  • These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
  • These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.
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Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

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It would take about 4.8 years to travel from earth to Saturn.

<h3>How long would it take?</h3>

We know that speed is expressed as the ratio of distance to time. In this case, we are trying to know ow many years would it take to reach the planet Saturn travelling at 21 thousand miles per hour.

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= 4.8 years

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