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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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Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a
Naily [24]

Answer:

The angular acceleration of the pencil<em> α  = 17 rad·s⁻²</em>

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

   mg(L / 2)sin(10) = (1/3 m L²)(α) 

   α(f) = 3/2(g) / Lsin(10)

   α  = 3/2(9.8) / 0.150sin(10)

  <em> α  = 17 rad·s⁻²</em>

Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>

3 0
3 years ago
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0
3 years ago
A marketing executive is investigating whether this year’s advertising campaign has resulted in greater mean sales compared with
djverab [1.8K]

We make use of a one-sample t-test for a population mean.

One-sample t-test for a population mean

Option B

<h3> Sample mean and Sample standard deviation</h3>

A Sample Standard Deviation is the root-mean square of the  data  minus the sample mean,

The sample mean is is the mean of the randomly selected sample

Therefore, For a data or sample where we have no information on population standard deviation and here only one sample group is compared, we make use of a one-sample t-test for a population mean

A one-sample t-test for a population mean

More on Probability

brainly.com/question/795909

7 0
3 years ago
Its important please answer ​
Yuri [45]

Answer: Well we see that v= u- ft/m

lets say that ft= feet and m= mass

so you divide those and subtract u

leaving you with the value of v which is 13

7 0
3 years ago
Mer and count Assessment items Assessment started:
RoseWind [281]

Answer:

Asteroids

Explanation:

Asteroids are the small and large block of rocks that are found to be scattering space. There are numerous asteroids present in the asteroid belt which is located between the planet of Mars and Jupiter. These range from a  few meters to hundreds of kilometers. They move so rapidly, at about 80,000 km/hr and there occurs constant collision with one another. These asteroids sometimes enter into the atmosphere of earth, which can cause a great mass extinction event.

These are believed to be the remnants particles that are present in space right after the formation of the solar system.

5 0
3 years ago
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