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Lesechka [4]
3 years ago
6

Charlie drove around the block at constant velocity, is it true or false?​

Physics
1 answer:
Paha777 [63]3 years ago
7 0

Answer:

FALSE I THINK

Explanation:

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1. ¿Cuál es el actor clave de del exceso de peso corporal?
Slav-nsk [51]

El factor mas importante para el exceso de peso es un  exceso de energía creada por una alimentación excesiva

El peso de un cuerpo es definido por la relación entre la energía requerida para los procesos vitales del cuerpo, sus actividades físicas diarias y la energía suministra en forma de alimentos.

Cuando estos dos parámetros están en balance el peso es estable, pero cuando la cantidad de alimentos aumenta o el valor energético de los mismo aumenta se tiene un exceso de energía que el cuerpo almacena en forma de grasa corporal, este el el factor mas importante para el exceso de peso.

En conclusión el factor mas importante para el exceso de peso es un  exceso de energía por una alimentación excesiva

aprende mas acerca del peso corporal aquí:

brainly.com/question/13032223

8 0
2 years ago
Does the tide that the moon raises on the earth different?
alukav5142 [94]

Answer:

No the gravity of the moon pulls the water making high tide

Explanation:

7 0
2 years ago
A hollow conductor is positively charged. Asmall uncharged metal ball is lowered by a silk thread through asmall opening in the
dimulka [17.4K]

Answer:

Explanation:

According to the property of a conductor, the entire charge will reside on the outer surface of the conductor, there is no charge on the inner side of the conductor. As the uncharged metal ball touches the inner surface of the conductor, it does not attain any charge as the inner side of the conductor has no charge.

So option (c) is correct.

8 0
3 years ago
A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th
Pavel [41]

Answer:

The block's mass should be 3m

Explanation:

Given:

Cart with mass m

From the conservation of energy before mass is added,

  \frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}

Where A = amplitude of spring mass system, k = spring constant

  A = v\sqrt{\frac{m}{k} }

Now new mass M is added to the system,

   \frac{1}{2} (m +M ) v^{2}  = \frac{1}{2}  k A^{2}

  A = v \sqrt{\frac{m +M }{k} }

Here, given in question frequency is reduced to half so we can write,

   f' = \frac{f}{2}

Where f = frequency of system before mass is added, f' = frequency of system after mass is added.

        \omega ' = \frac{\omega}{2}

\sqrt{\frac{k}{m +M} }  = \frac{\sqrt{\frac{k}{m} } }{2}

   \frac{k}{m +M } = \frac{k}{4m}

   M = 3m

Therefore, the block's mass should be 3m

8 0
3 years ago
What do we call the small changes that
Lelu [443]

Answer:

The "butterfly Effect"

Explanation:

The "butterfly effect" will probably have big changes in the future.

6 0
2 years ago
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