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Rzqust [24]
4 years ago
6

What is the answer to this question?

Physics
1 answer:
dem82 [27]4 years ago
3 0
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.

There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:

(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2

We can reduce this to:

(15 x 10)m/36 s^2 = 150 m/36 s^2

Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
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A graduated cylinder is filled with water to the 24-mL mark. When a rock is placed in the cylinder, the level rises to the 53-mL
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Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π
Kruka [31]

Answer:

 ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w = \sqrt{\frac{g}{L} }

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

          T = 2\pi  \ \sqrt{\frac{L}{g} }

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = 2\pi  \ \sqrt{\frac{L + \Delta L}{g} }

            T = 2\pi  \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }

in general the length increments are small ΔL/L «1, let's use a series expansion

           \sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...  

we keep the linear term, let's substitute

           T = 2\pi  \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L}  )  

if we do

           T = T₀ + ΔT

           

           T₀ + ΔT = 2\pi  \sqrt{\frac{\Delta L}{g} }  + \pi  \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}

            T₀ + ΔT = T₀ + \pi  \sqrt{\frac{1}{Lg} } \ \Delta L

            ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

4 0
3 years ago
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