He answer is A. <span>encourage agricultural usage in the watershed
if you want to read it for yourself go to
www.nature.org/ourinitiatives/regions/northamerica/unitedstates/indiana/journeywithnature/watersheds...
hope this helps you!!</span>
The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

<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>






<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
The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.
Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.
The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.
To know more about working distance
brainly.com/question/13551539
#SPJ4
That's unaccelerated motion,
and constant velocity.
Answer:
Into the page (Answer B)
Explanation:
Recall that the magnetic field that is created around a current conducting wire follows the right hand rule, where the direction of your fingers point in the direction of the magnetic field vectors when your thumb points in the direction of the current flow.
Use this to see that as you place the thumb of your right hand on the page pointing to the right (direction of the current in the wire). As you do so, you find the four fingers of your right hand entering the paper.