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expeople1 [14]
3 years ago
11

Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in

icial y su aceleración
Chemistry
1 answer:
faust18 [17]3 years ago
3 0

Answer:

La velocidad inicial es 55 \frac{m}{s}y su aceleración es -10 \frac{m}{s^{2} }

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad  varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

x=x0+v0*t+\frac{1}{2} *a*t^{2}

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

  • x= 90 m
  • x0= 0 m
  • v0= ?
  • t= 2
  • a= ?

Reemplazando obtenes:

90=v0*2+\frac{1}{2} *a*2^{2}

90=v0*2+\frac{1}{2} *a*4

90=v0*2+2*a

Y por otro lado tenes:

  • x= 120 m
  • x0= 0
  • v0= ?
  • t= 3
  • a= ?

Reemplazando obtenes:

120=v0*3+\frac{1}{2} *a*3^{2}

120=v0*3+\frac{1}{2} *a*9

120=v0*3+\frac{9}{2} *a

Por lo que tenes el siguiente sistema de ecuaciones:

\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

v0= \frac{90-2*a}{2}

Reemplazando en la segunda ecuación:

120=\frac{90-2*a}{2} *3+\frac{9}{2} *a

Resolviendo:

120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a

120=135-3*a +\frac{9}{2} *a

120-135=-3*a +\frac{9}{2} *a

-15=\frac{3}{2} *a

\frac{-15}{\frac{3}{2} } =a

-10=a

Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

v0= \frac{90-2*(-10)}{2}

Resolviendo:

v0= \frac{90+20}{2}

v0= \frac{110}{2}

v0=55

<u><em>La velocidad inicial es 55 </em></u>\frac{m}{s}<u><em>y su aceleración es -10 </em></u>\frac{m}{s^{2} }<u><em></em></u>

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ELEN [110]

Answer:

The result is a superposition which is twice the amplitude of each input wave. Φ = π means the two waves are completely OUT OF PHASE, and so add completely destructively. The result is a superposition which has no amplitude at all.

Explanation:

The result is a superposition which is twice the amplitude of each input wave. Φ = π means the two waves are completely OUT OF PHASE, and so add completely destructively. The result is a superposition which has no amplitude at all.

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1 year ago
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__________ often cause overactive eye movement. A. Stimulants B. Depressants C. Barbiturates D. None of the above
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<u>Stimulants </u> often cause overactive eye movement

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3. What noble gas would be part of the electron configuration notation for Mn?
shusha [124]

Answer:

Argon {Ar}

Explanation:

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7 0
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Which formula equation represents the burning of sulfur to produce sulfur dioxide?
Leni [432]

Answer:

S(s) + O2(g) --> SO2(g)

Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

Explanation:

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Oxygen = O2

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So we have;

S(s) + O2(g) --> SO2(g)

The crrect option is option A. Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

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What mass of potassium chloride, KCl, is produced when 12.6 g of oxygen, 02, is produced?
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Mass of KCl= 19.57 g

<h3>Further explanation</h3>

Given

12.6 g of Oxygen

Required

mass of KCl

Solution

Reaction

2KClO3 ⇒ 2KCl + 3O2

mol O2 :

= mass : MW

= 12.6 : 32 g/mol

= 0.39375

From the equation, mol KCl :

= 2/3 x mol O2

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