A runner begins from rest and accelerates at a rate of 0.22 meters per second' for 15

Physics

1 answer:

ziro4ka [17]3 years ago

7 0

Answer:

25 m

Explanation:

Given:

v₀ = 0 m/s

a = 0.22 m/s²

t = 15 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (15 s) + ½ (0.22 m/s²) (15 s)²

Δx = 24.75 m

Rounded to two significant figures, the runner moves a distance of 25 m.

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Answer:

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$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is: