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anyanavicka [17]
3 years ago
6

La superficie de unas botas suman 400 cm2 y la persona que las usa tiene 45 kg de masa, calcula la presión que ejerce sobre el p

iso. NOTA: Se debe encontrar la fuerza, para lo cual, se debe obtener el peso utilizando la fórmula de masa por gravedad que es 9.8 m / s2
Physics
1 answer:
liubo4ka [24]3 years ago
4 0

Answer:

11025 N / m²

Explanation:

Los siguientes datos se obtuvieron de la pregunta:

Área (A) = 400 cm²

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Presión (P) =?

A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Fuerza (F) =.?

F = m × g

F = 45 × 9,8

F = 441 N

A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:

1 cm² = 0,0001 m²

Por lo tanto,

400 cm² = 400 cm² × 0,0001 m² / 1 cm²

400 cm² = 0,04 m²

Por tanto, 400 cm² equivalen a 0,04 m².

Finalmente, determinaremos la presión ejercida de la siguiente manera:

Área (A) = 0.04 m².

Fuerza (F) = 441 N

Presión (P) =?

P = F / A

P = 441 / 0,04

P = 11025 N / m²

Por tanto, la presión ejercida es 11025 M / m²

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Put the value of torque in the equation

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Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a
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Answer:

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Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

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Sum of the potential at point p is

V = Va + Vb

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V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]

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likewise,

\frac{1}{(b^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }

hence,

V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

1/r is common to both equation

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by binomial expansion of fractional powers

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if we expand the expression we have the equivalent as shown

{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )

also,

{(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )

the above equation becomes

V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]

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Answer

V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }

OR

V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }

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