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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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Answer:

F₁ = 4.29 x 10⁵ N

Explanation:

The total force required to move the freight train with the given acceleration is given by the following formula:

F = ma + f

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F = Total Force Required from both engines = ?

m = equivalent mass of system = 2(8 x 10⁵ kg) + 5.5 x 10⁵ kg = 21.5 x 10⁵ kg

a = required acceleration = 5 x 10⁻² m/s²

f = force of friction = 7.5 x 10⁵ N

Therefore,

F = (21.5 x 10⁵ kg)(0.05 m/s²) + 7.5 x 10⁵ N

F = 8.575 x 10⁵ N

Now, for identical forces in each engine can be given as:

Force exerted by each engine = F₁ = F/2

F₁ = 8.575 x 10⁵ N/2

<u>F₁ = 4.29 x 10⁵ N</u>

3 0
3 years ago
A water bug is suspended on the surface of a pond by surface tension (water does not wet the legs). The bug has six leg, and eac
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Answer:

m = 2.2 x 10⁻⁴ kg = 0.22 g

Explanation:

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Weight\ of bug\ per\ unit\ length = Surface\ Tension\ of\ Water\\\frac{mg}{L} = Surface\ Tension\ of Water\\m = \frac{(Surface\ Tension\ of\ Water)(L)}{g}

where,

m = mass of bug = ?

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L = 30 mm = 0.03 m

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m = \frac{(0.072\ N/m)(0.03\ m)}{9.81\ m/s^{2}} \\

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A 2 kg toy cart and a 6 kg toy cart have a spring compressed between them. When the spring expands, it sends the 2 kg toy cart o
harkovskaia [24]

Answer:

The speed of second toy cart is 4 m/s.

(c) is correct option

Explanation:

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Mass of first toy cart = 2 kg

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We need to calculate the speed of second toy cart

Using formula of momentum

m_{1}v_{1}=m_{2}v_{2}

Where, m₁ = mass of first toy cart

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