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2 years ago

Which of the following is NOT associated with Urban Sprawl?

Engineering

1 answer:

pochemuha2 years ago

8 0

The option that is not associated with the given term called urban sprawl is; Option A: Blocking high views

What is Urban Sprawl?

Urban sprawl is defined as the rapid expansion of the geographic boundaries of towns and cities which is often accompanied by low-density residential housing and increased reliance on the private automobilefor movement.

Looking at the given options, "blocking high views" is the option that is not typically a problem associated with urban sprawl because urbanization usually takes place on relatively flat levels.

The missing options are;

a. blocking high views

b. destroying animal habitats

c. overrunning farmland

d. reducing green space

Read more about urban sprawl at; brainly.com/question/504389

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It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc

ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength $(\sigma_{cd})=630\ MPa = 630*10^6\ Pa$ , Fracture strength

$(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa$

a) The critical length ( $L_c$ ) is given by:

$L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm$

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

$\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456$

6 0

3 years ago

When the compression process is non-quasi-equilibrium, the molecules before the piston face cannot escape fast enough, forming a

muminat

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

5 0

3 years ago

Read 2 more answers

You don't have to notify employees that a lockout/tagout is about to begin?

Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

3 0

3 years ago

A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th

Inessa [10]

Answer:

attached below

Explanation:

4 0

3 years ago

A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi

____ [38]

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

$V_{max}=\omega A$

now by putting the values

$V_{max}=\omega A$

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

6 0

3 years ago