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Rus_ich [418]
2 years ago
10

El medio ambiente y su relacion con la educacion fisica??

Physics
1 answer:
Whitepunk [10]2 years ago
3 0

Answer:

La Educación Ambiental es un proceso educativo continuo que persigue hacer sensible, formar y modificar actitudes de forma objetiva sobre la realidad global del medio, tanto natural como social, con este trabajo consideramos que la asignatura de Educación Física en sí,

Explanation:

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The condition of a country’s depends on its people’s ability to exchange money for goods and services.
vaieri [72.5K]

Answer;

-Economy

The condition of a country’s economy depends on its people’s ability to exchange money for goods and services.

Explanation;

Economy is the state of a country or region in terms of the production and consumption of goods and services and the supply of money.

An economy encompasses all activity related to production, consumption and trade of goods and services in an area. An economy applies to everyone from individuals to entities such as corporations and governments.

There are four different types of economies; traditional economy, market economy, command economy and mixed economy. Each type of economy has it’s own strengths and weaknesses.


6 0
3 years ago
Read 2 more answers
The scientific theories to the law of parsimony?
Anna007 [38]

Answer:

vfx local KFC kid lsd Ltd Ltd KFC

4 0
3 years ago
A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.90 x 105 P
worty [1.4K]

Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

Inside pressure P_{in}= 1.90\times10^{5}\ Pa

Area A=3.90\times10^{-4}\ m^2

Outside pressure = 1 atm

We need to calculate the magnitude of the force

Using formula of force

F_{net}=F_{in}-F_{out}

F_{net}=(P_{in}-P_{out})A

Where, P_{in} =inside Pressure

P_{out} =outside Pressure

A = area

Put the value into the formula

F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}

F_{net}=34.59\ N

Hence, The magnitude of the force is 34.59 N.

6 0
3 years ago
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to
Degger [83]

The acceleration of the  box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

  • m is the mass of the box
  • a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

  • θ is the inclination of the plane
  • μ is coefficient of friction

F(net) = 170 -  (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the  box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

#SPJ4

8 0
1 year ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
2 years ago
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