Elastic potential energy stored in a spring is
(1/2) · (spring constant) · (stretch or compress)² .
PE = (1/2) · (100 N/m) · (0.1 m)²
PE = (50 N/m) · (0.01 m²)
PE = (50 · 0.01) (N · m / m²)
PE = 0.5 N · m
PE = 0.5 Joule
6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12
Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
![W = \dfrac{k}{2}[x^2]_6^{10.1}](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Bx%5E2%5D_6%5E%7B10.1%7D)
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.
Answer:The thymus stores and produces T lymphocytes and also releases hormones which aid in the maturation of the T lymphocytes while the spleen stores and release lymphocytes as well as macrophages which eradicate foreign substances that may harm the body.
Explanation:
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
