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Anit [1.1K]
3 years ago
15

Determine how much concrete you will need for a slab which is 50 feet by 30 feet wide and 1 foot thick

Engineering
1 answer:
Levart [38]3 years ago
6 0

Answer:

Amount of concrete need to make slab = 1,500 feet³

Explanation:

Given:

Length of slab = 50 feet

Width of slab = 30 feet

Height of slab = 1 feet

Find:

Amount of concrete need to make slab

Computation;

Amount of concrete need to make slab = Volume of cuboid

Volume of cuboid = (l)(b)(h)

Amount of concrete need to make slab = (50)(30)(1)

Amount of concrete need to make slab = 1,500 feet³

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2.11 Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface
alexandr402 [8]

Answer:

The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.

The cost associated with the heat loss through the windows for an 8-hour flight is:

For aerogel windows: $17.472 (most efficient)

For polycarbonate windows: $262.08

For soda-lime glass windows: $1,238.016 (least efficient)

Explanation:

To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:

\frac{dQ}{dt}=\dot Q =-kS\nabla \vec{T}

In this case:

\dot Q =k\frac{S}{L} \Delta T

If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).

To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).

To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):

Cost=C_{hc}\cdot \dot Q \cdot t \cdot n=1\frac{\$}{Kwh} \cdot \dot Q \cdot 8h \cdot 130

6 0
3 years ago
What have you learned about designing solutions? How does this apply to engineering? Think of some engineering solutions that st
Andrew [12]

Answer:

In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.

Explanation:

4 0
3 years ago
An office worker claims that a cup of cold coffee on his table warmed up to 80°C by picking up energy from the surrounding air,
kherson [118]

Answer:

The claim is false and violate the zeroth law of thermodynamics.

Explanation:

Zeroth law of thermodynamics refers to thermal equilibrium among  elements. It states that  elements which different temperatures will reach the same temperature at the endgame if they are close enough to interact each other. This temperaure is called <em>equilibrium temperature and it is always a intermediate value between the element with highest temperature and the element with the lowest one. So there is no way </em> a cup of cold coffee on a table can warm up to 80°C picking up energy from the surrounding air at 25°C because the cup can only reach a temperature closer to the surrounding air temperature which will be the equilimbrium temperature for that case.

4 0
3 years ago
Read 2 more answers
Which Battery should be removed first to prevent short circuits?
Cerrena [4.2K]
You are going to have to be more specific than that, perhaps add a picture next time ? Unfortunately, this question can’t be answered.
8 0
3 years ago
El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par
Lesechka [4]

Answer:

La probabilidad pedida es 0.717

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

X: '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria X será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

X ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ =\frac{1}{E(X)}

En donde E(X) es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ =\frac{1}{E(X)} ⇒ λ =\frac{1}{600}

Si reemplazamos el valor de λ en (I) obtenemos :

X ~ ε (\frac{1}{600})

La función de distribución de X (por ser una variable aleatoria exponencial) es :

F_{X}(x)=P(X\leq x)=  1 - e ^ ( - λx) con x > 0 y F_{X}(x)=0 en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de X obtenemos :

F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}  

Dado que la variable aleatoria X se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

P(X>200)    

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717

7 0
3 years ago
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