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Eddi Din [679]
1 year ago
7

you are about to work some magic with matter. you have a rectangular cube of matter in its solid state. it measures 3 in x 4 in

x 5 in. you have available a rectangular container measuring 5 in x 6 in x 3 in. you also have a cylindrical container witb base radius of 2.5 in and height 8 in. illustrate how that matter "fits" in each container when it is in its solid, liquid, and gaseous state.
Physics
1 answer:
Elenna [48]1 year ago
6 0

In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.

The matter in gaseous state can be expanded to occupy the volumes of the container.

<h3>Volume of each of the container</h3>

The volume of each of the container is calculated as follows;

<h3>Volume of the rectangular container</h3>

V = 5 in x 6 in x 3 in

V = 90 in³

<h3>Volume of the cylindrical container</h3>

V = πr²h

V = (π)(2.5 in)²(8 in)

V = 157.1 in³

<h3>Volume of the matter</h3>

Vm = 3 in x 4 in x 5 in

Vm = 60 in³

<h3>Matter in solid and liquid state</h3>

Matter has fixed volume in solid and liquid state.

In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.

<h3>Matter in gaseous state</h3>

Matter has no definite volume in gaseous state.

The matter in gaseous state can be expanded to occupy the volumes of the container.

Learn more about states of matter here:

#SPJ1

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Answer:

a) La magnitud del desplazamiento es de 5 m

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c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}

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\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}

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\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

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c) La dirección de la fuerza viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right )  = 38.9^{\circ}

La dirección del plano viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right )  = 38.9^{\circ}

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

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