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goldenfox [79]
4 years ago
6

Reto: Aníbal desea construir una carpa de base hexagonal de 1,40 m de altura. El vértice de la base es de 0,80m y la longitud de

la arista lateral mide 1,80m. ¿cuanto de tela necesito para cubrir la superficie lateral de la carpa? ¿cuanto de tela necesito para cubrir la base de la pirámide, que es un hexágono regular? Cuál es el volumen de la pirámide? ¿cuál es el área total de la pirámide?
Physics
1 answer:
Rudiy274 years ago
3 0

Answer:

1.- para cubrir la superficie lateral 4.32 metros²

2.- Area de la base  2.15 metros²

3.- Volumen 1 m³

4.- Area total 6.47 metros²

Explanation:

El área lateral sera calcular el area de seis triangulos iguales cuya base es

0.80 de base x 6 lados = 4.80 metros perimetro de la base

4.80 perimetro de base x 1.80 arista lateral / 2 = 4.32m2

Area de la base:

Perímetro x Apotema / 2

Siendo la Apotema la altura de los triangulos que componen un hexagono calculada utilizando el teorema de pitágoras:

\sqrt{0.8^2 + (0.8/2)^2} = Apotema

Apotema = 0.894427191

Area: 4.80 x 0.894427191 / 2 = 2.146625258

sumando el area de la base mas el area lateral se obtiene el area total

2.15 + 4.32 = 6.47 metros

Volumen de la pirámide:

Area de la Base x Altura / 3

2.15 x 1.40 / 3 = 1.00333 m3

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Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

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<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

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From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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