Answer:
c
Step-by-step explanation:
X²+20=0
x=±√(-80)/2
x=±4i√5/2=2i√5
factorised it is then
(x+2i√5)*(x-2i√5)
= 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:
35
Step-by-step explanation:
Base times height divided by 1/2 or 2 and 10 times 7 = 70 and 70 ÷ 2= 35
A+b 5
------- = ---
2a-b 4
b 5
-------- = ---- => 9b = 5a+45 => b = (5a+45)/9
a+9 9
a+b 5
------- = --- => 4a+4b = 10a-5b
2a-b 4
9b = 6a
9(5a+45) =6a divide both sides by 3
3(5a+45) = 2a
15a +135 = 2a
13a = -135
a = -135/13 = -10,38
a= -10,38
9b=6a
3b=2a
3b = 2(-10,38)
3b = -20,76
b = -20,76/3
b = -6,92
hope this helped
