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rewona [7]
3 years ago
12

¿Qué juegos se practicaban en esa Época? ¿Qué nombre tenían los juegos? ¿En qué consistía cada uno de los juegos? ¿Cómo se inici

aba y finalizaba el juego?
Physics
1 answer:
ivolga24 [154]3 years ago
7 0

La respuesta correcta para esta pregunta abierta es la siguiente.

A pesar de que se te olvidó especificar el país al que te refieres y a la fecha específica para saber de qué época estás hablando, te podemos ayudar comentando lo siguiente.

¿Qué juegos se practicaban en esa Época?

Anterior al surgimiento de la era digital y la era de los "gammers," los niños y la juventud en general salían a la calle o al parque a realizar diferentes actividades físicas y recreativas. Había juegos de ronda, juegos deportivos, juegos de destreza, y juegos de mesa.

¿Qué nombre tenían los juegos?

Había juegos que se llamaba "Doña Blanca," "La Roña," "Lobo estás ahí," "Las escondidas," "canicas," el yoyo," "el trompo," "el burro castigado," "las cebollitas," "el látigo," además de las "cascaritas" que eran juegos de soccer o futbol americano en las calles y con equipos formados por tus amigos.

¿En qué consistía cada uno de los juegos?

Si jugabas canicas, tenías que golpear una canica con otra para desplazarla y llevarla a cierto lugar. O tenías que meterla en agujeros.

Si jugabas, "Doña Blanca," formabas un círculo entrelazado de las manos. Mientras cantabas la canción de Doña Blanca, alguien que estaba afuera del círculo trataba de romperlo, tratando de soltar algunas de las manos entrelazadas.

La verdad era sumamente divertido, creativo, te reías, mucho, y lo mejor de todo era que hacías ejercicio y no te quedabas sentado todo el día frente a una pantalla de computadora.

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Does a calculation lose any information when converted from centimeters to meters? Explain
german

Answer:

No

Explanation:

There are many units that are used to measure length of an object. For example centimeters, meters, millimeters etc.

There is a relationship between any of two units to measure lengths. If we want to convert some length from cm to m, it can be done as follows :

1 cm = 0.01 m

or

1 m = 100 cm

When we use this conversion, the calculation remains the same. Only the way to represent it will be different.

Hence, there is no lose of information when converted from centimeters to meters.

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4 years ago
Assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single
Free_Kalibri [48]
May be b I’m not for sure tho
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3 years ago
Water moves through a constricted pipe in steady, ideal flow. At the
Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

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<em>L</em><em>ike the alto in a choir the octavina plays the lower melody . what musical element does this tell?</em>

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