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sladkih [1.3K]
2 years ago
6

The increase of kinetic energy at the square of the speed of your vehicle has a major influence on all motor vehicles in three p

articular situations. What are they
Physics
1 answer:
Viktor [21]2 years ago
5 0

This implies that stopping distance and impact force grow as a function of speed. The best ways to improve manoeuvrability and lessen crash severity are to drive at an appropriate pace and to slow down as soon as you spot dangers in front of you.

Keep in mind that stopping distance increases with speed; at 50 mph, it is four times longer than at 25 mph, and at 75 mph, the force of impact is nine times greater.

<h3>What is the impact of speed on kinetic energy ?</h3>

When your car expends or absorbs energy to speed up or slow down, you may feel a pull or a jolt, called impulse. Impulse increases as the energy or force increases, and increases as the duration of the force decreases. You'll feel a harder jolt if you speed up or slow down suddenly.

  • Consider: coming to a stop from 60 mph in ten seconds doesn't hurt you or your vehicle because the force of this event is spread out over a long time. But if you hit a wall and come to a stop in just half a second, you'll feel twenty times the impulse, causing severe damage.

Learn more about Kinetic energy here:

brainly.com/question/25959744

#SPJ4

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Un tubo metálico tiene 100 metros de longitud cuando está a 0 oC y 100,13 metros cuando se calienta hasta 100 oC. ¿Cuál es el co
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Answer:

 α= 1.3 10-5 ºC⁻¹

Explanation:

La dilatación termica de los cuerpos esta dada por la relación  

      ΔL = L₀ α ( T -T₀)

en este caso nos piden el coeficiente de dilatación térmica

     α =DL/L₀ DT

calculemos

      α = (  100,13 -100)/[100 (100 – 0)]

       α = 1,3 10-5 ºC⁻¹

Traduction

The thermal expansion of bodies is given by the relationship

       ΔL = L₀ α (T -T₀)

in this case they ask us for the coefficient of thermal expansion

        α = ΔL / L₀ ΔT

    let's calculate

        α = (100,13 -100) / [100 (100 - 0)]

        α= 1.3 10-5 ºC⁻¹

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