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3 years ago

What are the functions of each computer program

1 answer:

4 0

A function is a block of organized, reusable code that is used to perform a single, related action. Functions provide better modularity for your application and a high degree of code reusing. ... Different programming languages name them differently, for example, functions, methods, sub-routines, procedures, etc.

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Carbon dioxide flows at a rate of 1.5 ft3 /s from a 3-in. pipe in which the pressure and temperature are 20 psi (gage) and 120 °

Monica [59]

Answer:

the absolute pressure in the smaller pipe = 19.63 psi

Explanation:

Let A be the diameter of the first pipe = 3 inches

Let B be the diameter of the second pipe. = 1.5 inches

To feet (ft) ; we have

Diameter of the pipe A $D_1 = (\dfrac{3}{12})ft = 0.25 \ ft$

Diameter of pipe B $D_1 = (\dfrac{1.5}{12})ft = 0.125 \ ft$

Temperature T = 120° F = (120+ 460)°R

= 580 ° R

The pressure gage to atmospheric pressure ; we have:

$P_{Absolute }=P _{Atm} + P_{guage}$

where;

atmospheric pressure = 1.47 psi

pressure gage = 20 psi

$P_{Absolute }=(1.47+20)psi$

$P_{Absolute }=34.7 \ psi$

To lb/ft²; we have:

$P_{Absolute }=(34.7 *144 ) lb/ft^2$

$P_{Absolute }=$ 4.998.6 fb/ft²

The density of carbon dioxide can be calculated by using the relation

$\rho = \dfrac{P}{RT}$

$\rho = \dfrac{4996.8}{(1130 \ lb /slug ^0 R)*(580{^0} R)}$

$\rho = 7.64*10^{-3}\ slug /ft^3$

Formula for calculating cross sectional area is

$A = \dfrac{\pi}{4}D$

For diameter of pipe $D_1 = 0.025$

A₁ = $\dfrac{\pi}{4}*0.25^2$

A₁ = 0.04909 ft²

For diameter of pipe $D_2 - 0.0125$

A₂ $=\dfrac{\pi}{4}*0.125^2$

A₂ = 0.012227 ft²

Using the continuity equation to determine the velocities V₁ and V₂ respectively.

For V₁

Q = A₁V₁

V₁ = Q₁/ A₁

V₁ = 1.5/0.04909

V₁ = 30.557 ft/s

For V₂

Q = A₂V₂

V₂= Q₂/ A₂

V₂ = 1.5/0.04909

V₂ = 30.557 ft/s

Finally; using Bernoulli's Equation to the flow of the carbon dioxide from the larger pipe to the smaller pipe ; we have:

$p_1 + \dfrac{\rho V_1^2}{2}+\gamma Z_1= p_2 + \dfrac{\rho V_2^2}{2}+\gamma Z_2$

Since the pipe is horizontal then;

$\gamma Z_1= \gamma Z_2$

So;

$p_1 + \dfrac{\rho V_1^2}{2}= p_2 + \dfrac{\rho V_2^2}{2}$

$p_2 =p_1 +\dfrac{1}{2} \rho(V_1^2-V_2^2)$

$p_2 =4996.8+\dfrac{1}{2} *7.624*10^{-3}(30.557^2-122.23^2)$

$p_2 =4943.41 \ lb/ft^2$

To psi;

$p_2 =\dfrac{4943.41 }{144}psi$

$p_2 =34.33 \ psi$ gage

The absolute pressure in the smaller pipe can be calculated as:

$p_2 _{absolute} = 34.33 - 14.7$

$p_2 _{absolute} = 19.63 \ \ absolute$

Hence, the absolute pressure in the smaller pipe = 19.63 psi

7 0

3 years ago

According to information found in an old hydraulies book, the energy loss per unit weight of fluid flowing through a nozzle conn

nekit [7.7K]

Answer:

Yes equation is valid.

Explanation:

Given:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

Using SI units to assign dimensions to every quantity as follows:

Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]

Hose diameter D = [m]

Nozzle tip diameter d = [m]

Fluid velocity in the hose V = [ m s^-1 ]

Acceleration of gravity g = [ m s^-2 ]

Using the Given Equation and plug the SI units of respective quantities:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]

Simplify the equation above:

[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]

[m] = [m]

Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.

3 0

3 years ago

In addition to passing an ASE certification test, automotive technicians must have __________ year(s) of on the job training or

Lunna [17]

Two years or one year

8 0

2 years ago

The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes

kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0

3 years ago

Just need someone to talk to pls dont just use me for points

Allushta [10]

Answer:

well what do you wanna talk about friend?

Explanation:

7 0

3 years ago