All categories

2 years ago

Which of these processes uses a die and a press to form parts?

1 answer:

4 0

The process of die forming that uses a die and a press to form parts from the given processes is called; C: Hydroforming

<h3>What are some of the steps in the die forming operations?</h3>

There are different types of die forming in sheet metal operations. Now, Hydroforming is a specialized type of die forming that uses a high pressure hydraulic fluid to press room temperature working material into a die.

Tailor Welding is a process of making welded blanks made from individual sheets of steel of different thickness, strength and even coating that are joined together by the use of laser welding.

Read more about Welding operations at; brainly.com/question/9450571

You might be interested in

An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of

uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

$\Delta L=L\alpha\Delta T$

Now by putting the values

$\Delta L=750\times \25.5\times 10^{-6}\times 200$

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

3 0

3 years ago

Using leftover paint colors is acceptable in a paint shop and will help cut down on waste.

Luden [163]

Im pretty sure it is true !

3 0

2 years ago

Read 2 more answers

Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a numb

tekilochka [14]

Answer:

// Program is written in C++

// Comments are used to explain some lines

// Only the required function is written. The main method is excluded.

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

// The next line declares the final result of summation of divisors. The variable declared is also

//initialised to 0

int result = 0;

// find all numbers which divide 'num'

for (int i=2; i<=(num/2); i++)

{

// if 'i' is divisor of 'num'

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

cout<<result+1;

}

6 0

3 years ago

A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?

Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0

3 years ago

For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$