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2 years ago

Why do birds have wings

Engineering

2 answers:

scoundrel [369]2 years ago

6 0

Answer:

Their feathers are light and the shape of their wings is perfect for catching the air. Their lungs are great at getting oxygen and very efficient, so they can fly for very long distances without getting tired. They eat lots of high-energy food.

Elodia [21]2 years ago

4 0

Answer:

to fly and to slap

Explanation:

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A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa.Paddle-wheel work is done on the system until the pressu

AleksandrR [38]

Answer:

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

3 0

3 years ago

The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

5 0

3 years ago

Which type of work is a textile engineer most likely to do?

pickupchik [31]

Answer:

B because as a textile engineer, your job is to help design and create fabric, including the equipment and materials needed for fabrication.

7 0

3 years ago

A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig

jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

Section: 9-ft section of pipe.

Sum of forces perpendicular to member AC = 0

F_d - 0.8*W*L = 0

F_d = 0.8*10*9 = 72 lb

Sum of forces perpendicular to member BC = 0

F_e - 0.6*W*L = 0

F_e = 0.6*10*9 = 54 lb

F_d = 72 lb , F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

Section: Entire Frame.

Sum of moments about point B = 0

-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

-A_y*(1.5625) + 15 + 39.375 = 0

A_y = 34.8 lb

Sum of forces in vertical direction = 0

A_y + B_y - 0.8*F_d - 0.6*F_e = 0

B_y = 0.8*(72) + 0.6*(54) - 34.8

B_y = 55.2 lb

Sum of forces in horizontal direction = 0

A_x + B_x - 0.6*F_d + 0.8*F_e = 0

A_x + B_x = 0

Section: Member AC

Sum of moments about point C = 0

F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

72*2.5 - 34.8*12 - 9*A_x = 0

A_x = -237.6 / 9 = - 26.4 lb

B_x = - A_x = 26.4 lb

A_x = -26.4 lb , B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

Member AC:

From point A till just before point D

-0.6*A_x*x - A_y*0.8*x + M = 0

15.84*x - 27.84*x + M = 0

M = 12*x ..... max value at D, x = 12.25 in

M_max = 12*12.25/12 = 12.25 lb-ft

Member BC:

From point B till just before point E

-0.8*B_x*x + B_y*0.6*x + M = 0

-21.12*x + 33.12*x + M = 0

M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in

M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

5 0

3 years ago

Fluid fills the space between two parallel plates. The differential equation that describes instantaneous fluid velocity for uns

USPshnik [31]

Answer:

u/v = S (y²w) / m sinwt + y/h

Explanation:

see attached image

3 0

3 years ago