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2 years ago

What must engineers keep in mind so that their solutions will be appropriate?

Engineering

1 answer:

vekshin12 years ago

7 0

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

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Millions of years ago, the Sierra Nevada region began to be uplifted along a crack in Earth's crust. The region on the other sid

Evgen [1.6K]

Answer: The correct answer is : Fault block mountain with rough edges and steep cliffs

Explanation: Snowy saws are an example of a mountain chain blocked by faults. The snowy mountains were formed because the tectonic movement forced some segments of the earth's crust up into irregular pieces and others down.

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3 years ago

A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai

lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

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2 years ago

Airbags will deploy in a head on collision but not in a collision that occurs from angle

Aneli [31]

Answer:

Airbags will deploy in almost any angle.

Explanation:

Cars have sensors around them, so when the car gets hit, the sensors detect a crash and deploy the airbags to keep you safe.

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2 years ago

Each of the following activities are commonly performed during the implementation of the Database Life Cycle (DBLC). Fill in the

kicyunya [14]

Yessiree I agree with yu cause yu are right

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3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$