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3 years ago

What is the definition of insert view and why do we use it

Engineering

1 answer:

maw [93]3 years ago

4 0

Answer:

its a view point for auto cad

Explanation:

from my knowlege in IED we learned about it as a way of sing how an object would look in inventor or auto CAD

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2) The switch in the circuit below has been closed a long time. At t=0, it is opened.

saul85 [17]

Answer:

il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

il(t) = e^(-t/.01)

il(t) = e^(-100t) . . . amperes

8 0

2 years ago

Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra

VashaNatasha [74]

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

int *p;

}arr[1000];

2) struct array{

int *p;

}arr[1000];

6 0

3 years ago

A horizontal pipe has an abrupt expansion from D1 5 8 cm to D2 5 16 cm. The water velocity in the smaller section is 10 m/s and

anyanavicka [17]

Answer: Explanation: Application of the bernoulli's equation comes in from conservation of mass flow. The cross sectional area of the two pipes are calculated. from A = πD²/4 The velocity of water from conservation of mass flow is also calculated ; V2 = Ac1V1/Ac2 The Loss coefficient is then calculated from KL = (1 - Ac1²/Ac2²)² Then the head Loss (hL) is calculated The detailed calculated and appropriate steps is as shown in the attached files.

5 0

3 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$