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3 years ago

12

deep-sea trench under the ocean is MOST similar to a _____ on land. A. plateau or plain B. valley or canyon C. peak or mountain

D. forest or grassland

1 answer:

IceJOKER [234]3 years ago

5 0

Answer:

The answer is B, as they run deep, severely lower than the surrounding land.

I hope this helps!

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Water flows near a flat surface and some measurements of the water velocity, u, parallel to the surface, at different heights y

BabaBlast [244]

Answer:

a) since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b) The velocity according to the equation at y=0 is equal to 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary, this value is wrong hence the equation is NOT CORRECT

Explanation:

Given that;

range ⇒ 0 < y < 0

1ft is given by the equation u = 0.81 + 9.2y + (4.1 × 10³y³)

so u=velocity of water at different layers

y= height of the layer

a)

consider BG system of units

u(ft/s) = 0.81 + 9.2y + (4.1 × 10³y³)

and consider y=0.05 ft

u = 0.81 + 9.2(0.5) + (4.1 × 10³(0.5³)

u = 0.81 + 0.46 + 0.5125

u = 1.7825 ft/s lets say this is equation 1

now consider the SI system units

u(m/s) = 0.81 + 9.2y + (4.1 × 10³y³)

also consider y=0.05ft

1ft = 3.048×10⁻¹ (from conversion table)

so 0.05ft = 0.01524m

we substitute

u(m/s) = 0.81 + 9.2(0.01524m) + (4.1 × 10³(0.01524m)³)

u = 0.81 + 0.1402 + 1.4512×10⁻²

u = 0.9647 m/s

1m/s = 3.281 ft per seconds ( conversion table)

so

0.9647 m/s = 0.9647(3.281)

u = 3.165 ft/s lets say this is equation 2

now since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b)

we know that the velocity of water at the surface contact is zero

u=0

so from the equation

u = 0.81 + 9.2y + (4.1 × 10³y³)

at y = 0

u = 0.81 + 9.2(0) + (4.1 × 10³(0)³)

u = 0.81 ft/s

The velocity according to the above equation at y=0 is 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary this value is wrong hence the equation is NOT CORRECT

5 0

4 years ago

Hey how is everybody?

Tomtit [17]

Answer:

ig good

mostly bored

u would rlly drive that instead a ferrari??

Explanation:

7 0

3 years ago

Read 2 more answers

If in Example 1.2,q = (10 - 10e2) mC, find the current at t = 0.5 s.

NeTakaya

Answer:

pls mark me as brainliest

Explanation:

Answer: 7.36 mA

3 0

3 years ago

1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per

KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0

3 years ago

There are little to no benefits to supplementary field identification programs

VashaNatasha [74]

Answer:

True, supplementary field identification programs tend to limit the use of routine programs that target service delivery using routine systems.

Explanation:

When supplementary field identification programs are applied in a study, they have damaging effects to other systems and programs already in progress targeting certain/similar variables in a study group.Such programs are initiated to boost the already existing systems of programs that are in continuous application( routine basis). As a supplement , we expect more positive results in the rates per the variables included in a study.However, results has proved the opposite.For example, supplementary immunization activities applied in programs targeting demographic and health systems services reveled that such programs reduce the probability of receiving the services provided by other routine health systems conducting continuous vaccination programs to the target groups.

4 0

4 years ago