Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of

Answer:

Temperature T = 394.38 K

Explanation:

The full solution and detailed explanation regarding the above question and its specified conditions can be found below in the accompanying document. I trust my explanation will assist you in grasping this particular topic.

Response:

a) 4 kg/s

b) 99.61 °C

Rationale:

Refer to the pictures provided.

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

where D refers to diameter, L refers to length and f refers to feed while V represents speed

Tc =

Tc =

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 ×

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

Response:

The solution to this question is 1273885.3 ∅

Clarification:

The first step is to ascertain the required hydraulic flow rate liquid based on the working pressure if a cylinder with a piston diameter of 100 mm is utilized.

Given that,

The distance = 50mm

The time t =10 seconds

The force F = 10kN

The piston diameter = 100mm

The pressure = F/A

10 * 10^3/Δ/Δ

P = 1273885.3503 pa

Subsequently

Power = work/time = Force * distance /time

= 10 * 1000 * 0.050/10

which amounts to =50 watt

Power =∅ΔP

50 = 1273885.3 ∅

Answer:

Explanation:

Un dato importante: la división entera elimina la parte fraccionaria. Por lo tanto, el promedio de 8, 10, 5 y 4 se presenta como 6, no 6.75.

Observación: Las pruebas incluyen cuatro valores de entrada muy grandes cuyo producto causa desbordamiento. No necesitas hacer nada especial, solo ten en cuenta que la salida no representa el producto correcto (en realidad, el resultado de cuatro números positivos es negativo; sorprendente).

Envía lo anterior para evaluación. Tu programa no pasará las últimas pruebas (eso es normal) hasta que completes la segunda parte a continuación.

Parte 2: Además, se debe calcular e imprimir el producto y promedio usando aritmética de punto flotante.

Presenta cada número de punto flotante con tres dígitos después del punto decimal, lo cual puedes hacer así: System.out.printf("%.3f", tuValor);

Ejemplo: Si la entrada es 8, 10, 5, 4, la salida sería:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

double average_arith = (num1 + num2 + num3 + num4) / 4.0;

double product_arith = num1 * num2 * num3 * num4;

int result1 = (int) average_arith;

int result2 = (int) product_arith;

System.out.printf("%d %d\n", result2, result1);

System.out.printf("%.3f %.3f\n", product_arith, average_arith);

}

}

Expected output: 1600.000 6.750