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Trava [24]
2 years ago
9

Which type of engineers were the designers of the Great Pyramids of Egypt and the Great Wall of China?

Engineering
2 answers:
Gemiola [76]2 years ago
8 0
I think it’s structural engineers but still check with the others
Digiron [165]2 years ago
4 0

Answer: it's B Structural Engineers

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Why did you choose agricultural and biosystem as a course? Help me guys
Mazyrski [523]

Answer:

To be able to develop as a human and give back to the community. To ensure that people have safe food and water to drink, clean fuel and energy sources, and in general a safe enviornment to live in.

Explanation:

7 0
2 years ago
Which clauses in the Software Engineering Code of Ethics are upheld by a whistleblower (check all that apply). a. "Respect confi
garri49 [273]

Answer:

c. and d

Explanation:

As a whistle-blower, one of your aim is to guide against unethical dealings of other people , hence you are creating an environment that uphold ethical conduct,

In addition, whistle-blowing will disclose all imminent dangers to the software community thereby preventing security breaches.

6 0
2 years ago
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
natima [27]

Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0
2 years ago
A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp
Furkat [3]

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

5 0
3 years ago
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th
ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

7 0
3 years ago
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