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Helga [31]
3 years ago
12

A car moving at 11.6 m/s begins to accelerate at 5.22 m/s/s and reaches a final velocity of 31.5 m/s. How far did it travel duri

ng this acceleration?
Physics
1 answer:
mr_godi [17]3 years ago
8 0

Use the kinematic equation,

Vf^2 = Vi^2 + 2aX

31.5^2 = 11.6^2 + 2(5.22)(x)

Solve for x

Answer: 82.2m

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Answer:

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Explanation:

First, we calculate the spring constant of a single spring:

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Δx = extension = 4 cm = 0.04 m

Therefore,

k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\

For a load of 30 N, applying Hooke's Law:

\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\

Hence, the correct option is:

<u>B. 6 cm</u>

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