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Evgesh-ka [11]
2 years ago
6

Las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si

el cable tensor más corto mide 10m, ¿cuál es altura de un cable que se encuentra a 100m de distancia del centro?
Engineering
1 answer:
Sveta_85 [38]2 years ago
8 0

La altura es de 169.4 metros.

Dado que las dos torres que sostienen un puente colgante tienen una separación de 240m y una altura de 110m a partir de la carretera, si el cable tensor más corto mide 10m, para determinar cuál es altura de un cable que se encuentra a 100m de distancia del centro se debe realizar los siguientes cálculos, aplicando la ecuación parabólica:  

  • (240)² = 4P x (110-10)
  • 57600 = 4P x 100
  • 57600 = 400P
  • 57600/400 = P
  • 144 = P
  • 200 x 200 = 4 x 144 x (Altura - 100)
  • 40000 = 576Altura - 57600
  • 40000 + 57600 / 576 = Altura
  • 169.4 metros = Altura

Por lo tanto, la altura es de 169.4 metros.

Aprende más en brainly.com/question/20333463

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Answer:

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Explanation:

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An ideal Otto cycle models the behavior of an explosion engine. This cycle consists of six steps, as indicated in the figure. Prove that the performance of this cycle is given by the expression

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Diesel cycle:

The ideal Diesel cycle is distinguished from the ideal Otto in the combustion phase, which in the Otto cycle is assumed at a constant volume and in the Diesel at constant pressure. Therefore the performance is different.

If we write the performance of a Diesel cycle in the form

:

η = 1 - (1/r^(ρ-1)) × ( (r^ρ)-1)/(ρ^(r-1)) )

we see that the efficiency of a Diesel cycle differs from that of an Otto cycle by the factor in parentheses. This factor is always greater than the unit, therefore, for the same compression reasons r

                   diesel performance is less than otto performance

\ eta_ \ mathrm {Diesel} <\ eta_ \ mathrm {Otto} \,

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3 years ago
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Which of the following is not true about manufacturing employment in the U.S?
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3 years ago
The drag force, Fd, imposed by the surrounding air on a
Ad libitum [116K]

Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

the power in hp

<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

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     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

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      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

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3 years ago
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