Ask question

Ask question

All categories

3 years ago

5

THEME: What is the impact of technology on architecture?

1 answer:

abruzzese [7]3 years ago

8 0

Answer:

With increased technological knowledge and consequent decreased factors of ignorance, the structures have less inert masses and therefore less need for such decoration. This is the reason why the modern buildings are plainer and depend upon precision of outline and perfection of finish for their architectural effect.

You might be interested in

Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2

tresset_1 [31]

1.i am superman

2.175 175 175

3.em dilisues

4.bye classmate magbabalik pa ako

8 0

3 years ago

Water flows at a uniform velocity of 3 m/s into a nozzle that reduces the diameter from 10 cm to 2 cm. Calculate the water’s vel

katrin2010 [14]

Answer:

See the pictures attached

Explanation:

4 0

3 years ago

A set of civil construction plans shows a distance of 131 meters across the front of a piece of property. You need to convert th

Hoochie [10]

Answer:

429.5 ft

Explanation:

We are told that;

1 ft = 0.305 meters

We want to convert 131 meters to ft, by proportion we have;

Distance = (131 × 1)/0.305

Distance = 429.5082 ft

We want to approximate to the nearest tenth of a foot.

This gives us;

Distance = 429.5 ft

7 0

3 years ago

Click this link to view O*NET’s Work Activities section for Graphic Designers. Note that common activities are listed toward the

bixtya [17]

I can’t click on the link sorry man

8 0

3 years ago

Read 2 more answers

Technicians have to determine the flow rate of water in a pipe with the aid of a venturi installation and a mercury au.oaeter. T

Schach [20]

Answer:

Manometric difference x=142.85 mm.

Explanation:

Given :

Pipe diameter $d_1=40 mm$

venturi meter $d_2=20 mm$

We can know that discharge through venturi meter is given as

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}$

$A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}$

$Q=A_1V_1$

$Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s$

$0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}$

h=1.8 m

We know that $h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )$

Where x is the manometric deflection

⇒ $1.8=x\left (\dfrac{13600}{1000}-1\right )$

So x=14.28 mm

Manometric difference x=142.85 mm.

7 0

3 years ago