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babunello [35]
3 years ago
12

To be able to write an ss-domain equation for a circuit, use partial fraction decomposition to separate the terms in this equati

on, and then inverse transform the equation back into the time-domain. Using the mesh-current method to analyze the following circuit yields a set of two integro-differential equations which would be difficult to solve analytically. However, applying a Laplace transform allows for these equations to be rewritten as polynomial equations which can be manipulated algebraically.

Engineering
1 answer:
Ann [662]3 years ago
4 0

Answer:

Assuming it is for the circuit attached below. To construct a domain equivalent for t > 0

Explanation:

See hand written solution

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Which statement about lean manufacturing is true when you compare it to mass production?
Len [333]
Where are the statements then bbs lol
6 0
3 years ago
What is a ton of refrigeration?
AURORKA [14]

Explanation:

The unit refrigeration is generally is given in terms of tons.In refrigeration compressor consume some amount of work to produce the cooling effect  with the help of evaporator and condenser.

In the simple words ton is the cooling load of refrigeration system.

So  

1 ton = 3.5 KW

1 ton = 12,000 BTU/hr

 

6 0
3 years ago
The Micro:bit is considered a what?
Scrat [10]

Sorry need points I'm new

3 0
3 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a
Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

8 0
3 years ago
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0
3 years ago
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