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4 years ago

Human settlement has 5 elements. How do you observe those 5 elements around you. Give examples.

Engineering

1 answer:

djverab [1.8K]4 years ago

3 0

Answer:

5 elements that human settlement required are discussed below in complete details.

Explanation:

Nature: The climate and the biosphere that give the basic requirements to survive.
Society: Agglomeration of characters that builds communication and connection administered by social artifacts such as political, social, cultural, and spiritual.
Man: makes a choice on what to do with the sources around them.
Networks: these are ways through which people, power, and knowledge flow.
Shells: Contain human movements and provide protection.

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A displacement transducer has the following specifications: Linearity error ± 0.25% reading Drift ± 0.05%/○C reading Sensitivity

White raven [17]

Answer:

The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

Explanation:

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

assume from specifications that k = 5v/5cm

= 1v/cm

u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2

= 0.01225v

v = 2v * 0.001

= 0.002v

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

= ((0.01225)^2 + (0.002)^2)^(1/2)

= 0.0124 cm

Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

8 0

3 years ago

Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4

Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0

3 years ago

Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say

dedylja [7]

The. Answer will be D

3 0

4 years ago

An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i

Harman [31]

Answer:2058.992KJ

Explanation:

Given data

Mass of object $\left ( m\right )$ =521kg

initial velocity $\left ( v_0\right )$ =90m/s

Final velocity $\left ( v\right )$ =14m/s

kinetic energy of body is given by= $\frac{1}{2}$ $m$ $v^{2}$

change in kinectic energy is given by substracting final kinetic energy from initial kinetic energy of body.

Change in kinetic energy= $\frac{1}{2}\times m$ $\left ( V_0^{2}-V^2\right )$

Change in kinetic energy= $\frac{1}{2}\times521$ $\left ( 90^{2}-14^2\right )$

Change in kinetic energy=2058.992KJ

7 0

3 years ago

ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in

Mama L [17]

Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

$d_{p} =mN_{p}$

Where

m = module = 5

Np = number of teeth of pinion = 26

$d_{p} =5*26=130mm$ = 0.13 m

The pitch line velocity is equal to:

$V_{t} =\frac{d_{p}*2*\pi *w_{p} }{120}$

Where

wp = speed of the pinion = 1800 rpm

$V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s$

The factor B is equal to:

$B=\frac{(12-Q_{v})^{2/3} }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396$

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

$K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832$

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

$P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}$

The bending stress is equal to:

$\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}} \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa$

4 0

3 years ago