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VladimirAG [237]
3 years ago
7

Which of the following color schemes is composed of hues next to eachother on the color wheel ?

Engineering
1 answer:
Softa [21]3 years ago
3 0
Analogous color schemes are composed of hues next to each other
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Various factors to be considered in deciding the factor of safety?
Eduardwww [97]

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

4 0
3 years ago
How do I cancel my subscription
Aleksandr [31]
Hey, it depends what you subscribed to, or what platform you are trying to unsubscribe from?
7 0
3 years ago
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g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
yKpoI14uk [10]

Explanation:

\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}

Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

5 0
3 years ago
In highways the far left lane is usually the _____
Ivan
Fastest


(Known as the fast lane)
8 0
3 years ago
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A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig
Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0
3 years ago
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