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Angelina_Jolie [31]

3 years ago

5

Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ c

ấu ở trạng thái cân bằng. Tại thời điểm khâu 1 ở vị trí υ1 = 900 hãy tính áp lực tại các khớp động tại B, C và A.

1 answer:

lesya [120]3 years ago

3 0

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ShoppingBay is an online auction service that requires several reports. Data for each auctioned item includes an ID number, item

daser333 [38]

Answer:

START

READ ID_Number

READ Item_description

READ length_of_auction_Days

READ minimum_required_bid

IF minimum_required_bid GREATER THAN 100

THEN

DISPLAY

Item Details are

Item Id : ID_Number

Item Description: Item_description

Length Action days: length_of_auction_Days

Minimum Required Bid: minimum_required_bid

END

Explanation:

5 0

3 years ago

A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg

Eva8 [605]

Answer:

$\dot{w}$ = -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

$\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}$

Given that tank is insulated so $\dot{Q}=0$ and no mass is leaving so

$\dot{m_e}=0$

$\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt$

$m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t$

Mass conservation $m_2-m_1=m_e-m_i$

$m_1,m_2$ is the initial and final mass in the system respectively.

Initially tank is evacuated so $m_1=0$

We know that for air $u=C_vT ,h=C_p T$ , $P_2v_2=m_2RT_2$

$m_2=0.42 kg$

So now putting values

$0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300$

$\dot{w}$ = -0.303 KW

3 0

3 years ago

A digital filter is given by the following difference equationy[n] = x[n] − x[n − 2] −1/4y[n − 2].(a) Find the transfer function

slega [8]

Answer:

$y(z) = x(z) - x(z) {z}^{ - 2} - \frac{1}{4} y(z) {z}^{ - 2} \\ y(z) + \frac{1}{4} y(z) {z}^{ - 2} = x(z) - x(z) {z}^{ - 2} \\ y(z) (1 + \frac{1}{4}{z}^{ - 2}) = x(z)(1 - {z}^{ - 2}) \\ h(z) = \frac{y(z)}{x(z)} = \frac{(1 + \frac{1}{4}{z}^{ - 2})}{(1 - {z}^{ - 2})}$

The rest is straightforward...

6 0

3 years ago

A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a

OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS A B C

Benzene 1 4.72583 1660.652 -1.461

Toluene 2 4.07827 1343.943 -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5% = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis for all components

So for Benzene ; p × y1=x1 × p1s ------let this be equation 1

for Toluene ; p × y2=x2 × p2s ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0

3 years ago

Prepare deflection angles for staking the curves in problem 2-2. In every example assume that Sta. P.I. = 37+53.21. Carry the re

weeeeeb [17]

Answer:

ya mom

Explanation:

4 0

3 years ago