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3 years ago

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

Engineering

2 answers:

Gemiola [76]3 years ago

8 0

I think it’s structural engineers but still check with the others

Digiron [165]3 years ago

4 0

Answer: it's B Structural Engineers

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Pick a subjectarea/field/topic that you are interested in. For each of the following Bonham- Carver uses of GIS give an example

Vanyuwa [196]

Answer:

I hope following attachment will help you a lot!

Explanation:

3 0

3 years ago

A gas mixture containing 3 moles CO2, 5 moles H2 and 1 mole water is undergoing the following reactions CO2+3H2 →cH3OH + H2O Dev

Ilya [14]

Aaaaaaaaaaaaaaaaaaaaaa

7 0

2 years ago

A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above

creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation $N_{o} = N_{i} * \frac{E_{f}-E_{i} }{KT}$

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, $N_{d} = 1.65*10^{16} - 10^{16} \\ = 6.5 * 10^{15} per cm cube$

5 0

3 years ago

Read 2 more answers

Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1

tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R

where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0

3 years ago

g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress

zvonat [6]

Answer:

a) $\mathbf{\sigma _ 1 = 4800 psi}$

$\mathbf{ \sigma _2 = 0}$

b) $\mathbf{\sigma _ 1 = 6000 psi}$

$\mathbf{ \sigma _2 = 3000 psi}$

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = 0$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 4800 psi}$

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = \frac{Pd}{4t}$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 6000 psi}$

$\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}$

$\mathbf{ \sigma _2 = 3000 psi}$

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0

3 years ago