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aleksandr82 [10.1K]
3 years ago
5

Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by

a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)

Engineering
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

Explanation:

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a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, dq/dt = 100 W

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, T_0 = 20^0 C = 273 + 20 = 293 K

Temperature of the inside of the box, T_{in} = ?

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

5 0
4 years ago
When a tensile specimen is stretched in the plastic region to an engineering strain of 0.2, calculate the amount of cold work pe
Valentin [98]

Answer:

0.2 x 100

Explanation:

Engineering strain is the original crossection/original crossection

cold work percentage is

original crossection/original crossection x 100

4 0
2 years ago
Consider an ideal gas undergoing a constant pressure process from state 1 to state
Radda [10]

Answer:

s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,

We obtain the expression to compute the specific entropy change:

s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})

Best regards.

6 0
4 years ago
6. At a construction site, cement, sand, and gravel are used to make concrete. The ratio of cement to sand to gravel is 1 to 2.4
S_A_V [24]

Answer:

Mass of cement used is 62.5 lb

Mass of gravel used is 225 lb

Explanation:

The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6

So, for 150 lb of sand

C : S : G = 1 : 2.4 : 3.6

\frac{C}{S}=\frac{1}{2.4}\\\Rightarrow C=S\frac{1}{2.4}\\\Rightarrow C=150\frac{1}{2.4}\\\Rightarrow C=62.5\ lb

Mass of cement used is 62.5 lb

\frac{S}{G}=\frac{2.4}{3.6}\\\Rightarrow G=S\frac{3.6}{2.4}\\\Rightarrow C=150\frac{3.6}{2.4}\\\Rightarrow C=225\ lb

Mass of gravel used is 225 lb

7 0
3 years ago
Fracture Mechanics: A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa m ( ) is exposed to a st
vampirchik [111]

Answer:

a)  927 MPa

b)The specimen will experience fracture

Explanation:

a) Calculate critical stress for brittle fracture

σ = fracture toughness / (y √ π * surface crack)

  = 45  /  ( 1  \sqrt{\pi * ( 0.75*10^-3)}  )

  = 927 MPa

b) since critical stress( 927 MPa)  < 1000 MPa

hence : The fracture will occur

7 0
3 years ago
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