Answer: 17.75p + 8 ≤ 150 OR p ≤ 8
Step-by-step explanation:
The equation that is equivalent is -2+(-7).
= 1/2kx2
U = potential energy of a spring at a certain position
k = the spring constant, specific to the spring, with units N/m.
x = distance the spring is stretched or compressed away from equilibrium
Potential Energy: Elastic Formula Questions:
1) A spring, which has a spring constant k = 7.50 N/m, has been stretched 0.40 m from its equilibrium position. What is the potential energy now stored in the spring?
Answer: The spring has been stretched x = 0.40 m from equilibrium. The potential energy can be found using the formula:
U = 1/2kx2
U = 1/2(7.50 N/m)(0.40 m)2
U = 0.60 N∙m
U = 0.60 J
Answer:
Umm 2 1/2
Step-by-step explanation:
im confident
