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LekaFEV [45]
3 years ago
8

Si se arroja una ròca en sentido horizontal desde un barranco de 100 m de altura. Choca

Physics
1 answer:
klasskru [66]3 years ago
7 0

x = 20 m

Calcular las componentes rectangulares de la velocidad inicial

En el lanzamiento horizontal la velocidad inicial vertical (Voy) es igual a cero, por lo que:

Vx = 20 m/s

Voy = 0

Paso No. 2: Anotar los datos para X y para Y. Recuerde que las velocidades y los desplazamientos

Para “X”Para “Y”Vx = 20 m/st =X =Voy = 0g= -9.81 m/s2Y = -5 m

Paso No. 3: Selección de las ecuaciones a utilizar

Recuerde que “X” que es la distancia horizontal que recorre un proyectil y para calcularla es necesario saber el valor de t (tiempo). Observe que en “Y ” tiene datos suficientes para calcular “t”. Paso 4: Resolver la ecuación considerando que Voy = 0, por lo que el primer término se anula.

Y= gt^2 / 2

Resolviendo para “ t “ :

t = 1.009637 s

Calculo de “ t “

 

Paso5: Calcular “ X “ utilizando la ecuación:  

Recuerde que “X” que es la distancia horizontal que recorre un proyectil y para calcularla es necesario saber el valor de t (tiempo). Observe que en “Y ” tiene datos suficientes para calcular “t”.

Resolviendo para “ X “ : X=Vx (t)

X = (20 m/s)(1.09637s)

X = 20 m

¡Por favor, marca Brainliest!

Gracias :)

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an object with a constant acceleration always have:

A. changing velocity

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2.04 s

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A ball is suspended by a lightweight string, as shown in the figure above.
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<u>Answer </u>

A. 1 and 2

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At 2 the potential energy is minimum and the kinetic energy is maximum.

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P.E = height × gravity × mass. The height referred here is the perpendicular height. Gravity and mass are constant in this case.

From the diagram it can be seen clearly that the vertical height from 2 to 1 is much greater than from 4 to 3.

This shows that the change in P.E is greater between 1 and 2 and so is kinetic energy.


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Alex swims at an average speed of 45m/min. how far does he swim in 1 min 24 sec?
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Answer:

63 m

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An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
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Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

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Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

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  F = qvB  

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   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

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