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2 years ago

1. ¿Cuál es el actor clave de del exceso de peso corporal?

Physics

1 answer:

Slav-nsk [51]2 years ago

8 0

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

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A boat has a propulsion system that consists of a pump that sucks water at the bow and presses it on the stern. All tubes are 5

Phoenix [80]

Answer:

The propulsion force at the moment of departure, is 49 N

Explanation:

Given;

diameter of tubes = 5 cm = 0.05 m

volumetric flow rate, V = 50 L/s = 0.005 m³/s

density of water, ρ = 1000 Kg /m³

hydraulic / propulsion force, F = ρVg

where;

ρ is the density of the fluid (water)

V is the volumetric flow rate of water

g is acceleration due to gravity

propulsion force, F = ρVg

propulsion force, F = 1000 x 0.005 x 9.8

propulsion force, F = 49 N

Therefore, the propulsion force at the moment of departure, is 49 N

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3 years ago

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3 years ago

Read 2 more answers

Which one is correct and why?

Gelneren [198K]

'A' and 'C' are exactly the same circuit, except the voltmeter's terminals are flipped.

'A' is the correct way to hook everything up.

If you start at the positive terminal of the battery, and follow the flow of current through the circuit and around to the negative terminal, you're following the path where the voltage gets lower and lower and lower all the way.

So each time you come to any device in the circuit ... whether it's a resistor or a meter ... you would be hitting the positive side of it first, and then the voltage where you come out on the other side of it would be lower.

So the left side of the resistor is more positive, and the right side is more negative. The voltmeter is connected correctly in 'A', but it's backwards in 'C'. If you connect the voltmeter like in 'C' and turn things on, the voltmeter will try to go down from zero. You can't read the number on it, and It's possible that the voltmeter might be damaged.

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3 years ago