If you replace the text value in an associative dimension, the text value will not change when the

Add the following vector given in rectangular form and illustrated the process graphically A = 16+j12, B= 6+j10.4

MariettaO [177]

Answer:

A=16+j12…'B=6+j10.4

Explanation:

add the following vector given in

3 0

1 year ago

The southernmost rim inlet elevation on the topographical survey is

Lynna [10]

The southernmost rim inlet elevation on the topographical survey is approximately 4.5 feet below the benchmark on the north side of West 55th Street.

What is topographical survey?

A topographical survey, often known as a land survey or topographical land survey, is a type of survey that includes contours. Topographical land surveying determines the exact location and specifications of natural and man-made features on a piece of land. The survey is then turned into a suitable and thorough plan, which incorporates man-made characteristics such as boundaries, neighbouring buildings, sidewalks, and so on. Natural elements such as trees, ponds, and ground contours are also detected by the topographical survey. Having a clear and accurate map of your property might help you avoid costly downstream mistakes caused by unforeseen challenges. It can supply you with the information you need about the land before making any alterations to it.

To learn more about topographical survey

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6 0

1 year ago

It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0

3 years ago

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago

. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298

andriy [413]

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

5 0

3 years ago