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3 years ago

10

list out main types of material used in design and Manufacture of product give one example for each in engineering application ?

1 answer:

cluponka [151]3 years ago

7 0

One of the types of product design is interface design. Another type of product design is called visual product design.

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According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work

fiasKO [112]

Answer:

1090 Steel >1040 Steel > Pure aluminium >Diamond.

Explanation:

Toughness:

Toughness can be define as the are of load -deflection diagram up to fracture point.

Modulus of toughness can be defines as the area of stress-strain diagram up to fracture point.Modulus of toughness is the property of material.

So the decreasing order of toughness can be given as follows

1090 Steel >1040 Steel > Pure aluminium >Diamond.

8 0

3 years ago

Which of the following elements of the CIA triad refers to maintaining and assuring the accuracy of data over its life-cycle?

kenny6666 [7]

Answer:

Integrity: involves maintaining and assuring the accuracy of data over its life-cycle

Explanation:

Confidentiality: This is a CIA triad designed to prevent sensitive information from reaching the wrong people, while making sure that the right people have access to it.

Integrity: This is a CIA triad that involves maintaining the consistency, accuracy, and trustworthiness of data over its entire life cycle.

Availability: This is a CIA triad that involves hardware repairs and maintaining a correctly functioning operating system environment that is free of software conflicts.

Authentication:This is a security control that is used to protect the system with regard to the CIA properties.

4 0

3 years ago

At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing

Inessa [10]

Answer:

vB = - 0.176 m/s (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)

⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)

⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒ vB = - 0.176 m/s (↓-)

The pic can show the question.

7 0

3 years ago

Read 2 more answers

A gas metal arc welder is also known as a _____ welder.<br> A) TIGB) GTAWC) GMAWD) Resistance spot

nignag [31]

Answer:

GMAW

Explanation:

It's literally the initials of that type of welding

7 0

3 years ago

1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt

Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = $\frac{width}{period}$ ...............1

duty cycle = 1.5 × $10^{-3}$

and applied voltage will be

applied voltage = duty energy × voltage ...........2

applied voltage = 1.5 × $10^{-3}$ × 2

applied voltage = 3 mV

so current will be

current = $\frac{applied\ voltage}{resistance}$ ................3

current = $\frac{3}{150}$

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = $\frac{1.5}{18.75*10^{-6}}$

battery life = 80000 hours

battery life in year = $\frac{80000}{8760}$

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

4 0

3 years ago