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Viktor [21]
3 years ago
8

Create an application in VB that allows the user to enter an object’s mass, velocity, and height and displays the object’s KE an

d PE. The application should have a two functions: KineticEnergy that accepts an object’s mass (kilograms) and velocity (meters/second) as arguments and PotentialEnergy that accepts an object’s mass (kilograms), acceleration due to gravity (m/s2 ), and height (meters) as arguments. The function should return the objects KE and PE (3 decimal places).
Engineering
1 answer:
yanalaym [24]3 years ago
3 0

Answer:

see explaination

Explanation:

Module VBModule

Function KineticEnergy(ByVal mass As Decimal, ByVal velocity As Decimal) As Decimal

Dim result As Decimal

result = 0.5*mass*velocity*velocity

KineticEnergy = result

End Function

Sub Main()

Dim mass = Console.ReadLine()

Dim velocity = Console.ReadLine()

Console.WriteLine(FormatNumber(CDbl(KineticEnergy(mass,velocity)), 3))

End Sub

End Module

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1. Under what conditions can soils be chemically stabilized?
marshall27 [118]

Answer:

All will be Explained below.

Explanation:

1) Under which Condition can a soil be chemically Stabilize.

Answer

a). Plasticity Index :A soil with a high value of plasticity Index is not good for various engineering projects. The introduction of line helps in reducing plasticity due cation exchange reaction.Pozzolanic reaction over time reduces plasticity and increase index strength due to the formation of calcium - silicate hydrate.

7 0
3 years ago
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
3 years ago
Easy POINTS computer genius help me plz
Norma-Jean [14]

Answer:

dedededededede

Explanation:

4 0
3 years ago
A Carnot cooler operates with COP = 11, whose ambient temperature is 300K. Determine the temperature at which the refrigerator a
SashulF [63]

Answer:

275 Kelvin

Explanation:

Coefficient of Performance=11

T_H=\text {Absolute Temperature of high temperature reservoir=300 K}

T_L=\text {Absolute Temperature of low temperature reservoir}

\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=\frac{T_L}{300-T_L}\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}

8 0
3 years ago
2) The switch in the circuit below has been closed a long time. At t=0, it is opened.
saul85 [17]

Answer:

  il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

  il(t) = e^(-t/.01)

  il(t) = e^(-100t) . . . amperes

8 0
3 years ago
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