1. Delta, is formed by constructive erosion.
Answer:
Explanation:
Bobbitt worm ( Eunice aphroditois). This segmented polychaete marine worm can attain lengths of 10 feet. It bristles...
Goliath beetle ( Goliathus species). African goliath beetle ( Goliathus giganteus ). Five species of goliath beetle...
atlas moth ( Attacus atlas). Stop and rest your eyes on this lovely...
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!

<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>







<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Elasticity
C) Acceleration is the rate of change of velocity of an object. Velocity is the speed and direction of an object so acceleration is used to describe the rate of change. I hope this helps!!
The only way of telling about dark energy is our observation of how the universe has been expanding. It basically works the opposite as gravity, pushing things away from it. Thus, the closest answer would be D. The shape of galaxies in cluster galaxies.