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Mathematics 12 Online
OpenStudy (anonymous):

the distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inches with 75 pounds attached, how far will it stretch with 65 attached? Round to the nearest tenth of an inch.

OpenStudy (anonymous):

if 11inches=75lbs then x=65lbs, solve the system of equations

OpenStudy (anonymous):

SO the answer is 10.9

OpenStudy (anonymous):

i guess, let me ask what the others think?

OpenStudy (unklerhaukus):

Hooks Law: \(F(x)=-kx\)

OpenStudy (unklerhaukus):

find k

OpenStudy (anonymous):

but we dont have k

OpenStudy (anonymous):

it should be given

OpenStudy (anonymous):

9.53inches

OpenStudy (unklerhaukus):

\[F(x)=ma[\text{kg}\cdot \text{m}\cdot \text s^{-2}]\]

OpenStudy (anonymous):

@UnkleRhaukus I highly doubt that this needs physics laws to be solved...

OpenStudy (unklerhaukus):

well it does .

OpenStudy (anonymous):

okay, so would you mind giving @Crazyrayback a step by step explanation of how to get the answer?

OpenStudy (unklerhaukus):

\[ma=-kx\] \[m_1=75[\text{Oz}]=34.0[\text{kg}];\qquad x_1=11[\text{in}]=0.280[\text m]\]\[m_2=65[\text{Oz}]=29.5[\text{kg}];\qquad x_2=?\]

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