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laila [671]
2 years ago
8

11. What are restrictions when building or completing a challenge?

Engineering
1 answer:
katrin [286]2 years ago
3 0

Explanation:

The minimum exterior open spaces around buildings that are 55 metres or more, should be 16 metres. On sides where no habitable rooms face, a minimum space of 9 metres shall be left for heights above 27 metres.

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A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
diamong [38]

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

3 0
3 years ago
Ethylene glycol, the ingredient in antifreeze, does not cause health problems because it is a clear liquid.
netineya [11]
B. false

Explanation: because it is
5 0
2 years ago
How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?
Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the  question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = \frac{q^2Nd^2*Xn^2}{6Eo} * f

also note that ; Pactive ∝ Nd2 (

tD = K . \frac{Vdd}{(Vdd - Vt )^2}  since K = constant

Hence : Nd ∝ rt

5 0
3 years ago
List and briefly describe two modern's materials needs
Angelina_Jolie [31]

Answer:

Modern and smart materials for making the products are improved by developing new materials and find new uses for the existing. As, modern industrialization society is increased demand and quality of the product.

Two modern's materials are:

Carbon Fiber: As, carbon fiber is a strong material and it is light in weight. Designers used it because it is five times strong as steel and two times as stiff. Carbon fiber is basically made out of very thin strands of carbon.

Fiber Optics: It is a new technology as, it is used as transparent solid to transmitted light signals.

4 0
3 years ago
Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate
Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy +  flow work

E = \dot m ( \frac{v^2}{2] +  zg + p\nu)

E = \dot m( \frac{v^2}{2] +  zg + p_{atm} \frac{1}{\rho})

57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})

solving for flow rate

\dot m = 99.977we know that [tex]\dot m  = \rho AV

\dot m  = 780 \frac{\pi}{4} D^2\times 16

solving for d

99.97 = 780 \times \frac{\pi}{4} D^2\times 16

d = 0.090 m

so radius = 0.045 m

3 0
3 years ago
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