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Karolina [17]
3 years ago
14

Describe the placement of the views in a multi view drawing

Engineering
1 answer:
Marianna [84]3 years ago
8 0

Answer:

like a mountain place thanks #careonlearning

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What is the difference between a cost and a benefit?
goblinko [34]

Answer:

A cost is something you have to give up or sacrifice and a benefit is something that is gained or is helpful.

Explanation:

In a cost-benefit analysis of a system, an engineer is to simply look at the requirements for the system and determine the costs to build the system, both in financial value and energy value.  Additionally, the engineer needs to determine the benefits that would come from choosing a particular path of cost.  If the benefits outweigh the cost for the project, then the solution is accepted.  Else, the cost outweighs the benefit and the solution is rejected.

6 0
3 years ago
we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw
gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0
3 years ago
The arrival rate at a parking lot is 6 veh.min. Vehicles start arriving at 6:00PM and when the queue reaches 36 vehicles, servic
seraphim [82]

Answer:

Departure rate = 7.65 vehicle/min

Explanation:

See the attached file for the calculation.

5 0
3 years ago
The basic concept of feedback control is that an error must exist before some corrective action can be made?
weqwewe [10]

Answer:

The correct answer is True.

Explanation:

The feedback control system implies that to make a feedback, there must first be an error, otherwise there will be nothing to correct.

This system works so that there is an output that is controlled through a signal.

This signal will be feedback and it will signal an error which will be detected by a controller that will allow entry into the system.

In basic words, this system processes signals, samples them in the form of an output, and re-enters them feedback to detect the error signal.  

7 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
4 years ago
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