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katrin [286]
2 years ago
10

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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??? True or false. An element can only exist as an individual atom.
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7 0
3 years ago
A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body
JulijaS [17]
We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
S=30 m + 45m+65m=140 m

while the total time of the motion is
t=5 s+7s+5 s=17 s

So, the average speed of the body is:
v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s
8 0
3 years ago
It's for #8....I know the answer is 293.2 ft I just don't know how to get it
Oksi-84 [34.3K]
Kinectic Energy=1/2(mass)(velocity)^2 so 1.2=1/2(.0012)(position/2) so it travels 4000 m. Not sure how it is 293.2 ft
7 0
2 years ago
The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it
Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

5 0
3 years ago
The Hawaiian hot spot sits below the Pacific plate. As the plate moves over the hot spot, a chain of volcanoes is formed. The Ea
Viktor [21]

Answer:

3.2048179721\times 10^{-9}\ m/s

Explanation:

Assuming that the pacific plate moved 178 km in 1.76 million years.

s = Distance = 178 km

t = Time taken = 1.76 million years

Speed is given by

v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{178000}{1.76\times 10^6\times 365.25\times 24\times 3600}\\\Rightarrow v=3.2048179721\times 10^{-9}\ m/s

The speed of the plate is 3.2048179721\times 10^{-9}\ m/s

6 0
2 years ago
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