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3 years ago

Discuss in detail the manners of interaction with opposite gender

Engineering

1 answer:

siniylev [52]3 years ago

6 0

Answer:

8 Tips on Better Communication with the Opposite Sex

Put emotions away. Ladies, this one is more aimed at us, for the most part. ...

Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .

Put yourself in their shoes. .

Listen. ...

Respond. ...

Actually communicate. ...

Be detailed. ...

Don't communicate too much.

Explanation:

You might be interested in

Supercharging is the process of (a) Supplying the intake of an engine with air at a density greater than the density of the surr

iVinArrow [24]

Answer:

a)supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere

Explanation:

Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere.

By doing this , it increases the power out put and increases the brake thermal efficiency of the engine.It also increases the volumetric efficiency of the engine.

So the our option a is right.

4 0

3 years ago

What is the instantaneous center of zero velocity? List two approaches for determining the is the instantaneous center of zero v

Lera25 [3.4K]

Explanation:

Instantaneous center:

It is the center about a body moves in planer motion.The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body.About this center every particle of a body rotates.

From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

$I_{12},I_{23},I_{31} all\ are\ co-linear.$

5 0

3 years ago

Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press

Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

Given data:

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0

2 years ago

Ball Throws

professor190 [17]

Jake because the more speed, the more kinetic energy which = force.

3 0

2 years ago

Air flows through a convergent-divergent duct with an inlet area of 5 cm² and an exit area of 3.8 cm². At the inlet section, the

Luda [366]

Answer:

The mass flow rate is 0.27 kg/s

The exit velocity is 76.1 m/s

The exit pressure is 695 KPa

Explanation:

Assuming the flow to be steady state and the behavior of air as an ideal gas.

The mass flow rate of the air is given as:

Mass Flow Rate = ρ x A1 x V1

where,

ρ = density of air

A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²

V1 = inlet velocity = 100 m/s

For density using general gas equation:

PV = nRT

PV = (m/M)RT

PM/RT = ρ

ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k

ρ = 7.11 kg/m³

Therefore,

Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)

Mass Flow Rate = 0.27 kg/s = 270 g/s

Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:

Mass Flow Rate = ρ x A2 x V2

where,

ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)

A2 = exit area = 5 cm² = 5 x 10^-4 m²

V2 = exit velocity = ?

Therefore:

0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2

V2 = 76.1 m/s

Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:

P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2

Since, both inlet and exit are at same temperature.

Therefore, h1 = h2, and those terms will cancel out.

P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²

P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²

P2 = P1 + (1/2) ρ (V1² - V2²)

P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]

P2 = 680000 Pa + 14962.25 Pa

P2 = 694962.25 Pa = 695 KPa

4 0

2 years ago