A negative charged particle of known magnitude q is placed in a uniform electric field of known strength E. What additional info
1 answer:
Answer:d
Explanation:
Electric field is a region around charged particle where charged particle experience a force .
Magnitude of Force experienced by charged particle is given by
where F=force experienced by charged particle
q=charge of particle
E=strength of electric field
If q and E are given then no additional information is needed to determine the magnitude of force.
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There are 3 forces acting on the stoplight:
• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward
• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right
The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that
∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0
We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to
2<em>T</em> sin(<em>θ</em>) = <em>W</em>
2 (1000 N) sin(<em>θ</em>) = 100 N
sin(<em>θ</em>) = 0.05
<em>θ</em> ≈ 2.87°
If <em>y</em> is the vertical distance between the stoplight and the ground, then
tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)
Solve for <em>y</em> :
tan(2.87°) = (15 m - <em>y</em>) / (100 m)
<em>y</em> = 15 m - (100 m) tan(2.87°)
<em>y</em> ≈ 9.99 m
