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3 years ago

Please help asap! I will give brainlist!

Engineering

2 answers:

dexar [7]3 years ago

7 0

Answer:

i did this this is b

Explanation:

DENIUS [597]3 years ago

7 0

I did it last year it should be b from my knowledge

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A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a colu

Temka [501]

Answer:

(a) 126.66 kN (b) 31.665 kN (c) 258.49 kN (d) 506.64 kN

Explanation:

Solution

Given

A HSS152.4 × 101.6 × 6.4 structural steel is used as a column

Actual length of the column , L= 6 m

The elasticity modules, E = 200 GPa

The factor of safety with respect to failing buckling . F.S =2

Geometric properties of structural steel shapes for, A HSS152.4 × 101.6 × 6.4

the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4

For

(a) If the end condition is pinned - pinned

The effective length factor, K =1

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)

= 253319.85N

= 253.32N

The maximum safe load , Pallow = 253.32 /2 = 126.66kN

hence, the maximum safe for the column is 126.66kN

(b)If the end condition is fixed free-free

the effective length factor, K= 2

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²

= 63329.96N

=63.33kN

The maximum safe load, P allow = Pcr/F.S

P allow = 63.33/2

31.665 kN

Therefore the maximum safe for the column is 31.665 kN

The effective length factor K =0.7

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.7 * 6 * 10 ^3)²

=516979.2 8N

=516.98 kN

The maximum safe load, P allow = Pcr/F.S

P allow = 516.98 kN/2

=258.49 kN

Therefore the maximum safe for the column is 258.49 kN

(d) If the end condition is fixed -fixed

The effective factor, K =0.5

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²

=1013279.4 N

=1013.28 kN

The maximum safe load, P allow = Pcr/F.S

P allow = 1013.28 / 2

= 506.64 kN

P allow = 506.64 kN

Therefore the maximum safe for the column is 506.64 kN

8 0

4 years ago

Air is to be compressed so that its volume is reduced to half of its original volume if the initial pressure is 200 kPa and the

Alex_Xolod [135]

Answer:

$P_{2} = 527.803\,kPa$

Explanation:

The politropic relationship for a isentropic process is:

$\frac{P_{2}}{P_{1}} = \left(\frac{V_{1}}{V_{2}} \right)^{\gamma}$

Where $\gamma$ is the ratio of specific heats

The final pressure is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}}\right)^{\gamma}$

$P_{2} = (200\,kPa)\cdot (2)^{1.4}$

$P_{2} = 527.803\,kPa$

7 0

3 years ago

Choose the best data type for each of the following so that any reasonable value is accommodated but no memory storage is wasted

stiks02 [169]

Answer:

Explanation:

Part (a):

Statement : The number of siblings you have

Suitable Data type : Byte

Typical Value : From -128 and up to 127

Explanation: Byte data type is the most suitable since it can covers minimum and maximum number of siblings one can have.

Part (b):

Statement : Your final grade in this class

Suitable Data type : Char

Typical Value : 1 byte

Explanation: Grades is in the form of alphabetical letter which is either A, B, C, D, F or E which can be stored in character data type.

Part (c):

Statement : Population of Earth

Suitable Data type : Long

Maximum Value : 9223372036854775807

Explanation: Long Data takes up to 8 bytes and can store up to 9223372036854775807 which can cater for more than 36 billion. The population of earth is only around 7 billion currently making Long data type the most suitable data type to store earth population.

Part (d):

Statement : Population of US Country

Suitable Data type : Integer

Typical Value :2147483647

Explanation: Integer data type takes up to 4 bytes and can store up to 2147483647 making it suitable to store U.S population.

Part (e):

Statement : The number of passengers on bus

Suitable Data type : Byte

Typical Value :From -128 up to 127

Explanation: The typical maximum number of passengers of a bus are only around 72. Byte data type is the most suitable since it can cater the number up to 127.

Part (f):

Statement : Player's score in a Scrabble game

Suitable Data type : Short

Typical Value : 32767

Explanation: The maximum point can be scored in the Scrabble game is only 830 therefore the most suitable data type for this case is the short data type.

Part (g):

Statement : One team's score in a Major League Baseball game

Suitable Data type : Byte

Typical Value : From -128 up to 127

Explanation: The maximum point can be scored in the Base ball game is only 49 therefore the most suitable data type for this case is the Byte data type since it can cater up to 127.

Part (h):

Statement : The year an historical event occurred

Suitable Data type : Short

Maximum Value: 32767

Explanation: The historic event year can be any number from 1 to 2020 therefore the most suitable data type is the short data type.

Part (i):

Statement : The number of legs on an animal

Suitable Data type : Short

Maximum Value: 32767

Explanation: The most number of legs found are 750 legs therefore the most suitable data type is the short data type which can cater up to 32767.

Part (j):

Statement : The Price of an automobile

Suitable Data type : Float

Maximum Value: 340282350

Explanation: The most expensive car is around 15 million therefore the most suitable data type is the float data type which can cater up to 340 million.

3 0

3 years ago

1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.

sammy [17]

Answer:

A. Yes

B. Yes

Explanation:

We want to evaluate the validity of the given assertions.

1. The first statement is true

The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.

Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side

We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.

For notation purposes;

We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles

a/Sin A = b/Sin B = c/Sin C

2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate

So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A

That would be;

a^2 = b^2 + c^2 -2bcCosA

So yes, the cosine rule can be used for the scenario above

3 0

3 years ago

Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$

Evaporation rate = $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

6 0

3 years ago