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miss Akunina [59]
3 years ago
14

Water exerts little pressure on a building so it has no implications on foundation design.

Engineering
1 answer:
777dan777 [17]3 years ago
7 0

Fact

Explanation:

<em>If </em><em>wrong </em><em>Im </em><em>sorry </em><em>Correct </em><em>me </em><em>In </em><em>The</em><em> </em><em>c</em><em>om</em><em>m</em><em>ent </em><em>please </em>

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Free brainlist because im new and i just want to but you have t friend me first
Amiraneli [1.4K]
Okay sure.









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4 0
3 years ago
A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value
Dominik [7]

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

m = \dfrac{22}{32}=0.6875\ s^2/ft

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

C_c=2\sqrt{0.6875\times 58.66}

Cc= 12.7 lb.sec/ft

5 0
3 years ago
Is it a problem that the stress intensity factor, K, from Irwin's near-tip approximation approaches infinity as you get close to
valina [46]

Answer:

No it is not a problem

Explanation:

It is not a problem because the stress intensity factor K would approach infinity as you get close to a crack tip and the intensity factor would approach Zero as you get too far away from the crack tip and this is simply because a crack is a notch with zero tip radius .

and The application of stress intensity factor k in respect to present fatigue crack tip is termed " linear elastic fracture mechanics "

3 0
3 years ago
A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
4 years ago
Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
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