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Inessa05 [86]
3 years ago
7

La base de los tema relacionados a las ciencia de las ingeniería es?

Engineering
1 answer:
Yanka [14]3 years ago
5 0

Answer:

La ciencia y la ingeniería conciben el mundo como comprensible, con reglas que gobiernan su funcionamiento y que a través de un estudio cuidadoso y sistemático se puede evidenciar mediante patrones consistentes que permitan la oportunidad de examinar las características fundamentales que mejor describen los fenómenos.

Explanation:

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Technician A says you should place the air ratchet setting to
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I think it’s b sry if I’m wrong tho
6 0
3 years ago
Name the seven physical qualities for which standards have been developed.
SIZIF [17.4K]

Answer:

HUMAN DEVELOPMENT

MOTOR BEHAVIOR

EXERCISE SCIENCE

MEASUREMENT AND EVALUATION

HISTORY AND PHILOSOPHY

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Explanation:

6 0
3 years ago
A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the
sergeinik [125]

Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = \frac{weight}{area}

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = \frac{5000}{4.45}

                        = 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

7 0
2 years ago
Ext
Galina-37 [17]
Engineering is the technical
8 0
2 years ago
Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.1
Tamiku [17]

Answer:

(a) the mass flow rate of air is 5.351 kg/s

(b) the input power required is 2090.786 kW

Explanation:

Given;

initial pressure, P₁ = 100 kPa

initial temperature, T₁ = 20 °C

Final pressure, P₂ = 1.8 MPa

Final temperature, T₂ = 400 °C

Inlet area of the compressor = 0.15 m²

outlet area of compressor = 0.078 m²

velocity of air = 30 m/s

Part (a) mass flow rate of air through the inlet

Mass flow rate = Area x velocity = density x volumetric rate

m = Av = ρV

from ideal gas law, PV = nRT and ρ = m/V

substitute these values in the above equations, we will have;

m = \frac{PAv}{RT}

m = \frac{100000*0.15*30}{287*(20+273)}= 5.351 \ kg/s

Part (b) the required power input

W + m(h_1+\frac{v_1{^2}}{2}) = mh_2

where;

W is the input power

m is the mass flow rate

h₁ is the initial enthalpy

h₂ is the final enthalpy

initial and final enthalpy are obtained from steam table using interpolation;

h₁ = 293.166 kJ

h₂ = 684.344 kJ

W + m(h_1+\frac{v_1{^2}}{2}) = mh_2\\\\W = mh_2 - m(h_1+\frac{v_1{^2}}{2})\\\\W = 5.351 ( 684.344) - 5.351 (293.166 + \frac{30^2}{2000}) \\\\W = 3661.925 \ kW -1571.139 \ kW\\\\W = 2090.786 \ kW

7 0
3 years ago
Read 2 more answers
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