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cricket20 [7]
3 years ago
5

What was the code phrase the japanese used to initiate the attack on pearl harbor

Engineering
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer:

Tora!

Explanation:

It was an expression that illustrated the complete surprise of the attack.

yan [13]3 years ago
6 0

Answer:Itaka

Explanation:

You might be interested in
Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu
aksik [14]

Answer:

Para x=0:

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-2.4 W/m^{2}  

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

\phi=-k\frac{dT}{dx}(1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

\frac{dT(x)}{dx}=300x-30

Remplezando esta derivada en (1):

\phi=-0.04(300x-30)

Para x=0:

\phi=0.04(30)

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-0.04(300*0.3-30)

\phi=-2.4 W/m^{2}    

Espero que te haya ayudado!

4 0
3 years ago
Multiple Choice
12345 [234]

Answer:https://global.oup.com/us/companion.websites/9780199385423/student/ch6/mcq/     just go here

Explanation:

6 0
3 years ago
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

4 0
3 years ago
What are the optical properties of steel
dezoksy [38]

Answer:

A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.

In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.

Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.

Explanation:

3 0
3 years ago
A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp
Nikolay [14]

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

6 0
3 years ago
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