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TRADICIÓN E INNOVACIÓN S.A. de C.V.<br />

CosA =<br />

Despejando h (hipotenusa), tenemos que:<br />

Despejando b (altura) tenemos:<br />

Volumen del recipiente con agua:<br />

Volumen del cono:<br />

308<br />

( c)(<br />

a)<br />

h<br />

( c)(<br />

a)<br />

=<br />

h =<br />

cos A<br />

0.<br />

47m<br />

h = = 0.<br />

477m<br />

cos<br />

( 10°<br />

)<br />

2<br />

h = a +<br />

b<br />

2<br />

2 ( 0.<br />

47m)<br />

= 0.<br />

m<br />

2<br />

b = h − a = 0.<br />

477 −<br />

3246<br />

0.<br />

3246m<br />

39.<br />

37in<br />

= 12.<br />

77in<br />

1m<br />

2<br />

3<br />

( )( 18.<br />

50in)<br />

( 43.<br />

51in)<br />

46,<br />

782<br />

V = π π<br />

=<br />

2*<br />

* r h =<br />

in<br />

46,<br />

782in<br />

3<br />

V<br />

2 ( 18.<br />

50)<br />

( 12.<br />

77in)<br />

1m<br />

39.<br />

37in<br />

π<br />

=<br />

3<br />

3<br />

2 ()() r h<br />

= 0.76m 3<br />

3<br />

( 1m)<br />

3<br />

3<br />

V = 4576.<br />

81in<br />

=<br />

=<br />

3<br />

π<br />

39.<br />

37in<br />

0.<br />

075m<br />

Volumen total del recipiente con agua es 1.25m 3 +0.075m 3 = 1.325m 3<br />

Calculando el área superficial expuesta a la presión atmosférica<br />

Área superficial del cilindro :<br />

As =<br />

π<br />

( )( D)(<br />

h)

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