2 years ago

Last summer, Brianna planted 2 plants

Mathematics

1 answer:

Volgvan2 years ago

4 0

Answer: how many square feet are there? multiple each square foot by 2

Step-by-step explanation:

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A force of 12 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i

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Answer:

Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:

Normal length of the spring = 8 in or $\frac{8}{12}$ ft

= $\frac{2}{3}$ ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= $\frac{11}{12}$ 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( $\frac{2}{3}$ )

k = 18

Work done (W) to stretch the spring by $\frac{11}{12}$ ft will be

W = $\int\limits^\frac{11}{12} _0 {kx} \, dx$

= $\int\limits^\frac{11}{12} _0 {(18x)} \, dx$

= $18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0$

= 9( $\frac{11}{12}$ )²

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720 pt/h=______ qt/min

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720 pt/hr = 6 qts/min

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