Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or
ft
=
ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
=
ft
Force applied to stretch the spring = 12 lb
By Hook's law,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k(
)
k = 18
Work done (W) to stretch the spring by
ft will be
W = 
= 
= ![18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0](https://tex.z-dn.net/?f=18%5B%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%5D%5E%7B%5Cfrac%7B11%7D%7B12%7D%7D_0)
= 9(
)²
= 7.56 lb-ft
Answer:
Evaluate: to determine or set the value or amount of; appraise: to evaluate the property. to judge or determine the significance, worth, or quality of; assess: to evaluate the results of an experiment. Mathematics. to determine or calculate the numerical value of (a formula, function, relation, etc.).
Hope this helps! Please mark Brainliest!
17500 i think but u didn’t give enough info so it could be any without the extra info
8(4x+6y-2z)
32x+48y-16z
Hope this helps!