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4 years ago

What historical event allowed both aerospace fields to make enormous strides forward? *

Engineering

1 answer:

galina1969 [7]4 years ago

6 0

Answer:

The world wars. Most notably world War II

Explanation:

The demand for aircrafts during these events led to extensive research into the design of aircrafts. Aircraft advanced within these years from a simple design to a more complex design; capable of carrying fire power and even became bomb equipped. Also, the material of choice of production moved from wood to metal and the engine was improved on to gain more speed and maneuverability.

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With thermodynamics, one cannot determine ________.

zloy xaker [14]

Answer:

Option B is correct.

Explanation:

With thermodynamics, one cannot determine the speed of the reaction.

Speed of the reaction is defined as the rate at which one component is changes to another component. With factor cannot be calculated from the thermodynamics branch of engineering.

With thermodynamics we can determine the value of the equilibrium constant,

the extent of a reaction, the temperature at which a reaction will be spontaneous & the direction of a spontaneous reaction. But we cannot determine the speed of a reaction.

Therefore option B is correct.

4 0

3 years ago

The reading on a mercury manometer at 70(°F) (open to the atmosphere at one end) is 25. 62(in). The local acceleration of gravit

Ilya [14]

The absolute pressure in psia being measured is; 27.228 psia

<h3>What is the absolute Pressure?</h3>

Formula for absolute Pressure is;

Absolute pressure = Atmospheric pressure + Gauge pressure

P_{abs} = P_{atm} + P_g

We are given;

P_atm = 29.86 (in Hg) = 14.666 psia

Density of mercury at 70 °F; ρ = 13.543 g/cm³

Mercury Manometer reading; h = 25.62 in

Acceleration due to gravity; g = 32.243 ft/s²

Gauge pressure of the mercury = ρgh = 13.543 * 25.62 * 32.243

When we multiply and covert to psia gives; P_g = 12.562 psia

Thus;

P_abs = 14.666 + 12.562

P_abs = 27.228 psia

Read more about Absolute Pressure at; brainly.com/question/17200230

7 0

2 years ago

A tungsten matrix with 20% porosity is infiltrated with silver. Assuming that the pores are interconnected, what is the density

daser333 [38]

Answer:

15.4 g/cm³, 17.4 g/cm³

Explanation:

The densities can be calculated using the formula below

ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)

fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8

a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³

b) density after infiltration with silver

fraction occupied by silver = 20 / 100 = 0.2

density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³

5 0

3 years ago

Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D

slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

6 0

3 years ago

Determine the maximum volume in gallons [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3

Olin [163]

Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

P=total pressure or absolute pressure=26psi=179213.28Pa

Patm= the atmospheric pressure =101325Pa

Poil=pressure due to the weight of olive oil=0.86αgh

α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

$h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m$ [/tex]

Now we can use the equation that defines the volume of a cylinder.

V= $V=\frac{\pi }{4} D^{2} h$

D=3ft=0.9144m

h=9.23m

solving

$V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3$

finally we use conversion factors to find the volume in gallons

$V=6.06m^3\frac{1000L}{1m^3} \frac{1gal}{3.785L} =1601gal$

3 0

4 years ago