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2 years ago

You accelerate from 2m/s to 6m/s while traveling a distance of 2m; what was your acceleration?

Physics

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

The acceleration is 8 m/s²

Explanation:

The given parameters are;

The initial velocity, u = 2 m/s

The final velocity, v = 6 m/s

The distance the acceleration took place, s = 2 m

The acceleration, a, can be found from the following kinematic equation;

v² = u² + 2·a·s

By substituting the values, we have;

6² = 2² + 2 × a × 2

6² - 2² = 2 × a × 2

32 = 4·a

a = 32/4 = 8 m/s²

The acceleration, a, of the given motion = 8 m/s².

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Answer:

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2 years ago

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

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v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

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The distance traveled is given by;

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Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

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