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AfilCa [17]
3 years ago
7

Pick a subjectarea/field/topic that you are interested in. For each of the following Bonham- Carver uses of GIS give an example

from your chosen subject (for example, wetland ecology) Organetation Visualization Spatial Query. Combination Analysis, Prediction

Engineering
1 answer:
Vanyuwa [196]3 years ago
3 0

Answer:

I hope following attachment will help you a lot!

Explanation:

You might be interested in
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
3 years ago
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
3 years ago
You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0
2 years ago
Please answer the questions !
gizmo_the_mogwai [7]

Answer:

120

Explanation:

6 0
3 years ago
Hi, any kind of help on these questions will be appreciated.
Zielflug [23.3K]

Answer:

IDK

Explanation:

8 0
3 years ago
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