Answer:
Approximately
(given that the magnitude of this charge is
.)
Explanation:
If a charge of magnitude
is placed in an electric field of magnitude
, the magnitude of the electrostatic force on that charge would be
.
The magnitude of this charge is
. Apply the unit conversion
:
.
An electric field of magnitude
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J
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<h3>Further explanation</h3>
Let's recall Elastic Potential Energy formula as follows:

where:
<em>Ep = elastic potential energy ( J )</em>
<em>k = spring constant ( N/m )</em>
<em>x = spring extension ( compression ) ( m )</em>
Let us now tackle the problem!
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<u>Given:</u>
mass of object = m = 1.25 kg
initial extension = x = 0.0275 m
final extension = x' = 0.0735 - 0.0275 = 0.0460 m
<u>Asked:</u>
kinetic energy = Ek = ?
<u>Solution:</u>
<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>





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<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>






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<h3>Learn more</h3>
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<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Elasticity
Because atoms is something that pops or has bubbles in it
Answer:
Secrets? Cuz if I share you the secrets then the secrets will no longer will be secrets.
It would be A:facing disagreement forces scientist to prove their theories more consistently :)