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2 years ago

How can I draw this image in 2D form

Engineering

1 answer:

Ket [755]2 years ago

5 0

Answer:

no it is not 2D

Explanation:

it is 3D

ok so follow these steps

- make hole

-make square

-make triangle

ok now your figure is ready

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What should be given to a customer before doing a repair?

natima [27]

A. I believe, lmk if I’m right

7 0

3 years ago

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0

3 years ago

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

$\sigma=\frac{F}{A}$

$F=\sigma*A$

Convert given values to base units:

275 MPa = $275*10^{6}$ Pa

325 $mm^{2}$ = 0.000325 $m^{2}$

Substituting in given values:

F = $(275*10^{6})*(0.000325)=89375$ N

3 0

2 years ago

Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa a

Elena L [17]

Answer:

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

$S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1})$

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

$S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50})$

$S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888)$

$S_2 - S_1 = 0.5864 - 0.6968$

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

6 0

2 years ago

A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. Duri

notsponge [240]

Answer:

LOS = A

Explanation:

Given all the parameters the level of service as seen from the attached graph

is LOS = A

<u>To determine the LOS from the attached graph </u>

<em>calculate the trial value of Vp</em>

Vp = V / PHF

= (100 + 150) / 0.95 = 263 pc/h

since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1

next we will calculate the flow rate

flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]

Fhr = 1 / 1.035 = 0.966 ≈ 1

next calculate the real value of Vp

Vp = V / ( PHF * N * Fhr * Fp )

= ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )

Vp ≈ 126 pc/h/In

Next calculate the density

D = Vp / S = 126 / ( 45 * 1.61 ) = 1.74 pc/km/In

3 0

2 years ago