We have that the workdone in stretching natural length to is mathematically given as
W=9/2lbft
<h3>Workdone in stretching natural length</h3>
Question Parameters:
- A force of 128 lb is required to hold a spring stretched 2 ft <em>beyond </em>its natural length.
- Stretching it from its natural length to 9 inches beyond its natural length
Generally the Hookes equation for the Force is mathematically given as
F=Kx
Where
3.2lb=2ft*k
Thereofore
k=16
Force becomes
f=kx
f=16x
Hence
![W=8[x^2]^{3/4}_{0}](https://tex.z-dn.net/?f=W%3D8%5Bx%5E2%5D%5E%7B3%2F4%7D_%7B0%7D)
W=9/2lbft
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Answer:75
Step-by-step explanation:First isolate x by getting rid of the 100 so the equation becomes 3x=x +150 then you subtract x From both sides so it’s 2x=150 then divide by two on both sides so you get X = 75
Answer:
63
Step-by-step explanation:
9(7+9) = 63 +81
Answer: 6x^2 - 9x + 3
Step-by-step explanation:
first, i multiplied the parentheses.
3x x 2x - 3x - 3 x 2x - 3 x (-1)
then, i calculated/multiplied it.
6x^2 - 3x - 6x + 3
finally, i collected the like terms.
6x^2 - 9x + 3