h =(3.7 - .58)m = 3.12m
Now put PE into KE and we have to use the formula:
√2gh (g = gravity and h = height) therefor:
√2 x 9.8 x 3.12
= 7.82m/s
I hope this helps!
Answer:
-67,500 kgm/s
Explanation:
1300 * 20 + 1100 * (-85) = -67,500 kgm/s
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
Answer:
Explanation:
a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m
So terminal speed V = 2 / 4.5
= .444 m /s
When it attains terminal speed , acceleration becomes zero
0 = g - B x .444
B = 22.25 s⁻¹
b ) At t = 0 , v = 0
a = g - B v
a = g at t = 0
c ) When v = .15
a = g - 22.25 x .15
= 9.8 - 3.31
= 6.5 m /s²
Answer:
if somthing is warm or if somthing moves it usally has energy