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777dan777 [17]
3 years ago
8

What does Clay say will happen if the system is rejected?

Engineering
2 answers:
juin [17]3 years ago
6 0

Answer:

the nation will suffer terrible consequences

Explanation:

I did that and got it right

Verdich [7]3 years ago
4 0

Answer:

C

Explanation:

The nation will suffer terrible consequences

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Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat
Len [333]

Answer:

my_mul:

.globl my_mul

my_mul:

   //Multiply X0 and X1

   //   Does not handle negative X1!

   //   Note : This is an in efficient way to multipy!

   SUB SP, SP, 16       //make room for X19 on the stack

   STUR X19, [SP, 0]    //push X19

   ADD X19, X1, XZR     //set X19 equal to X1

   ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

   CBZ X19, mult_eol

   ADD X9, X9, X0

   SUB X19, X19, 1

   B mult_loop

mult_eol:

   LDUR X19, [SP, 0]

   ADD X0, X9, XZR      // Move X9 to X0 to return

   ADD SP, SP, 16       // reset the stack

   BR X30

Explanation:

6 0
4 years ago
A metal having a cubic structure has a density of , an atomic weight of , and a lattice parameter of Å One atom is associated wi
Veronika [31]

Answer:

Explanation:

Answer: The crystal structure of the metal is BCC

Explanation:

we first calculate the volume of the unit cell.

Volume of unit cell= (a°)^3.

The lattice parameter here is a°.

Substitute (6.13 * 10^-8)cm for a°.

Volume of unit cell = (6.13 * 10^-8)^3 = 2.3034 * 10^-22 cm^3/cell.

To determine the crystal structure we use

Density (p) = {(Number of atoms per cell) (Atomic mass)} / {(volume of unit cell)(Avogrado constant)}.

Substitute 1.892g/cm^3 for p (6.02*10^23) atoms/mol for Avogrado constant 1.3921g/mol.

For atomic mass and (2.3034 * 10^-22) cm^3/cell for unit cell.

1.892g/cm^3 = {(Number of atoms per cell) (1.3291g/mol)} / {(2.3034 * 10^-22) (6.02 * 10^23 atoms/mol)}.

Changing the subject of formula we have :

Number of atoms per cell = {(2.3034 * 10^-22) * (6.02 * 10^23) * 1.892} / 132.91

Number of atoms per cell = 2.

Since the number of atoms per cell is 2, :. the crystal structure of metal is BCC.

Note: p = density

a° = a subscript o

4 0
3 years ago
When replacing a timing belt, many experts and vehicle manufacturers recommend that all of the following should be replaced exce
lora16 [44]

Answer:

Correct Answer:

A. water pump

Explanation:

<em>Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence.</em> Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.

<em>Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time. </em>

7 0
3 years ago
El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par
Lesechka [4]

Answer:

La probabilidad pedida es 0.717

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

X: '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria X será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

X ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ =\frac{1}{E(X)}

En donde E(X) es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ =\frac{1}{E(X)} ⇒ λ =\frac{1}{600}

Si reemplazamos el valor de λ en (I) obtenemos :

X ~ ε (\frac{1}{600})

La función de distribución de X (por ser una variable aleatoria exponencial) es :

F_{X}(x)=P(X\leq x)=  1 - e ^ ( - λx) con x > 0 y F_{X}(x)=0 en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de X obtenemos :

F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}  

Dado que la variable aleatoria X se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

P(X>200)    

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717

7 0
3 years ago
Suppose that we have a 1000 pF parallel-plate capacitor with air dielectric charged to 1000 V. The capacitors terminals are open
AlekseyPX

Answer:

A) initial stored energy = 5x10^-4 J

B) new voltage if space is doubled will be 2000V

C) if space is doubled, stored energy becomes 2x10^-3 J

D) the extra energy comes from the work done in moving the plates further apart

Explanation:

Detailed explanation and calculation is shown in the image below

8 0
4 years ago
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