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ivolga24 [154]
3 years ago
14

The magnitude of vector A is 2.5 m and is directed towards east. The magnitude and direction of A /2 will be​

Physics
1 answer:
const2013 [10]3 years ago
4 0

The magnitude and direction of half of the given vector A is 1.25 m east.

The given parameters;

  • <em>magnitude of vector A = 2.5 m east</em>

A vector can be described with both magnitude and direction. The magnitude of the vector is the scalar or numeric representation of the vector.

When the scalar representation has direction, we term the quantity a vector quantity.

The magnitude of half of the given vector is calculated as follows;

\frac{A}{2} = \frac{2.5 \ m}{2} = 1.25 \ m\ east

Thus, the magnitude and direction of half of the given vector A is 1.25 m east.

Learn more here:brainly.com/question/14586816

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Answer:

h=20.38 m

Explanation:

Given that

Initial speed of object u = 20 m/s

Acceleration  g= 9.8\ \frac{m}{s^2}

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Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
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Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

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En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

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