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tatiyna
3 years ago
11

One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that

time , the agency ends. In addition, the parties can specify that the agency is for a particular . Once that is achieved, the agency ends. Alternatively, the parties can include a specific event as a trigger for termination; once that event , the agency ends. The parties can terminate an agency relationship prior to any of the preceding events by agreement, or revocation by individual party. g
Engineering
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

Explanation:

Complete question:

Fill in the blanks

One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that time ,___1______the agency ends. In addition, the parties can specify that the agency is for a particular____2______ . Once that is achieved, the agency ends. Alternatively, the parties can include a specific event as a trigger for termination; once that event,_____3______ the agency ends. The parties can terminate an agency relationship prior to any of the preceding events by ______4_________agreement, or revocation_____5______ by individual party.

Answer

1) lapses

(2) purpose

(3) occurs / begins

(4) mutual

(5) either

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Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

 Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

  h=3374 KJ/Kg  ,s=6.59 KJ/Kg-K  (Super heated steam table)

<u>At 10 KPa:</u>

h_g=2675.1 KJ/Kg, h_f=417.51  KJ/Kg

s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= h_f+x(h_g- h_f)

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34  KJ/Kg

Lets take m is the mass flow rate of steam

So 5\times 10^3=m\times (3374-2449.34)

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0
3 years ago
If you could help it would mean alot.
scoundrel [369]

Answer:

D is the correct choice.

Explanation:

I'm assuming that this is probably a phase in the textbook or progarm you are studying, and this is just a matter of reading thoroughly.

Engineers usually benefit from catching a mistake, and would also benfit from keeping record of a misstep in order to remain clear of that mistake in the future.

Have a great day, and mark me brainliest if I am most helpful!

:)

8 0
3 years ago
Read 2 more answers
Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air con
PSYCHO15rus [73]

Answer:

See explaination

Explanation:

The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.

Kindly check attachment for the step by step solution of the given problem.

4 0
3 years ago
2. Can you make adjustments on a saw while the trigger is in the on position.
Liula [17]
Umm... Heck no, that is very dangerous and could cause you to loose a finger... OR A HAND. To be safe I would always turn it off but if the saw has an on and off switch and a rev switch I think it would be ok.
3 0
4 years ago
Read 2 more answers
Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB Originally the cable is unstretc
AnnZ [28]

Answer:

e_{ab} =  4.18*10^(-3)

Explanation:

This question will be solved with the help of diagram (see attachment)

Given:

Δ∅ = 0.50 degrees (correction)

BC = 800 mm

AC = 600 mm

Solution:

We use coordinate system with point C as origin (0,0)

Hence,

Point A = (600,0)

Point B = (0,800)

The change in length or displacement can be calculated BB' :

BB' = BC * tan (Δ∅) = (800 mm) * tan (0.5) = 6.9815 mm

Hence, Point B' = (-6.9815,800) and we calculate distance A and B

AB = \sqrt{600^2 + 800^2} = 1000 mm

We calculate distance A and B'

AB' = \sqrt{(600 - (-6.9815))^2 + (0-800)^2} = 1004.2 mm

Normal Strain in AB is:

e_{ab} = \frac{AB' - AB}{AB} \\\\= \frac{1004.2 - 1000}{100}\\\\= 4.18 * 10^(-3)

The solution is :

e_{ab} =  4.18 * 10^(-3)

6 0
3 years ago
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