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2 years ago

a low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a

1 answer:

5 0

A low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a powder-actuated tool.

<h3>What is a fastening system?</h3>

It should be noted that a fastening system is important to fixing rails to railroad ties.

In this case, a low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a powder-actuated tool.

Learn more about fasteners on:

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You might be interested in

A strain gauge with a 4 mm gauge length gives a displacement reading of 1.5 um. Calculate the stress at the location of the stra

Art [367]

Answer:

1) 75Mpa

2) 1.125 MPa

Explanation:

given data:

gauge length = 4 mm

displacement $= 1.5\mu m = 1.5\times 10^{-3} m$

a) structural steel

Young modulus for steel is $200 GPa = 200\times 10^3 MPa$

we know that

$E =\frac{stress}{strain}$

$Stress = 200\times 10^3 \times \frac{1.5\times 10^{-3}}{4}$

= 75Mpa

b) PMMA

Young's modulus $= 3GPa = 3\times10^3 MPa$

$stress = 3\times \frac{1.5\times10^{-3}}{4}$

stress = 1.125 MPa

3 0

3 years ago

Write a java program which will produce the following output. No variables are needed but the program must use numbers and math

Sav [38]

The question did not include the output to be produced. However, below are the outputs.

My age is 21

Twice my age is 42

Three times my age is 63

ANSWER:

public class NewClasss{

public static void main(String[] args) {

int x,y,z;

x = 21;

y = 2*x;

z=3*x;

System.out.println("My age is " + x);

System.out.println("twice my age is " + y);

System.out.println("three times my age is " + z);

}

8 0

3 years ago

The following figures were obtained in a standard tensile test on a specimen of low carbon steel with a circular sectional area:

liq [111]

Answer:

See Explaination

Explanation:

1)here for given stress strain curve graph is given as follows

where for getting stress,S=F/A=4F/(pi*(50*10^-3)^2)

for strain=e=dl/l=dl*10^-3/100 mm/mm or m/m

2)so graph is as follows

3)for getting youngs modulus of elasticity we must know slope of graph stress verses strain and for straight line in elastic region upto 12 point we have elastic region and from that we get E as

E=slope of graph for first 12 points=S/e=14.5665*10^9/.812=17.9390*10^9 N/m2

4)for getting ultimate tensile stress at which specimen bears maximum load without failure so we get UTS as

UTS=maximum load/area=40*10^6/1.9634=20.3728*10^6 N/m2

5)percentage reduction in area is given by

percentage reduction in area=[original area-final area/original area]*100

Percent reduction=[5062-10^2]*100/50^2=96%

6)percentage elongation is given by

percent elongation=[final length-original length/original length]*100

final length at fractureis=14.56+100=114.56 mm

so we get percent elongation as=[114.56-100/100]*100=14.56%

7)true fracture stress is given by load at fracture devided by true area at fracture

Sf=load/(true area)=4*28*10^3/(pi*(10*10^-3)^2)=356.5070*10^6 N/m2

8 0

3 years ago

In contouring, it is necessary to measure position and not velocity for feedback.

EastWind [94]

Answer:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.

(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.

ANSWER: a=> True.

(3). Job shop is another term for process layout.

ANSWER: b => False

JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.

(4). Airplanes are normally produced using group technology or cellular layout.

ANSWER: b => False.

(5). In manufacturing, value-creating time is greater than takt time.

ANSWER: a => True.

8 0

3 years ago

Consider a system having p processes, where each process needs a maximum of m instances of resource type R1. Given that there ar

Genrish500 [490]

Answer:

Consider a system consisting of 4 resources of same type that are share by 3 processes each of which needs at most two resources.Now we will show that the system is deadlock free.

If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done.

Consider a system with m resources of same type being shared by n processes. Resources can be requested and released by processes only on at a time. The system is deadlock free if and only if The sum of all max needs is < m+n .

We can understand the notion of a deadlock from the following simple real-life example.To be able to write a letter one needs a letter pad and a pen. Suppose there in one letterpad and one pen on a table with two persons seated around the table. We shall identify these two persons as Mr. A and Ms. B. Both Mr. A and Ms. B are desirous of writing a letter. So both try to acquire the resources they need. Suppose Mr. A was able to get the letter pad. In the meantime, Ms. B was able to grab the pen. Note that each of them has one of the two resources they need to proceed to write a letter. If they hold on to the resource they possess and await the release of the resource by the other, then neither of them can proceed. They are deadlocked. We can transcribe this example for processes seeking resources to proceed with their execution. Consider an example in which process P1 needs three resources r1 ; r2, and r3 before it can make any further progress. Similarly, process P2 needs two resources r2 and r3 Also, let us assume that these resources are such that once granted, the permission to use is not withdrawn till the processes release these resources. The processes proceed to acquire these resources. Suppose process P1 gets resources r1 and r3 and process P2 is able to get resource r2 only. Now we have a situation in which process P1 is waiting for process P2 to release r2 before it can proceed. Similarly, process P2is waiting for process P1 to release resource r3 before it can proceed. Clearly, this situation can be recognized as a deadlock condition as neither process P1 nor process P2 can make progress. Formally, a deadlock is a condition that may involve two or more processes in a state such that each is waiting for release of a resource which is currently held by some other process.

5 0

3 years ago