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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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What type of car engine is best for cold weather.
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Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

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Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

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Weight of water = 0.5 of weight of cement

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3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
max2010maxim [7]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

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P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

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We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

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Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

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(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

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Answer:

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