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Nitella [24]
4 years ago
12

A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring

constant of the spring system in the taptap?
Physics
1 answer:
ANEK [815]4 years ago
5 0

Answer:

29856.521 N/m

Explanation:

m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m

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Answer:

<em>The acceleration of the racecar is</em> \mathbf{11.17~m/s^2}

Explanation:

<u>Uniformly Accelerated Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Following the definition above, the acceleration is defined as:

\displaystyle a=\frac{v_f-v_o}{t}

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:

\displaystyle a=\frac{46.1-18.5}{2.47}

\displaystyle a=\frac{27.6}{2.47}

a = 11.17~m/s^2

The acceleration of the racecar is \mathbf{11.17~m/s^2}

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