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3 years ago

What is the theoretical density in g/cm3 for Lead [Pb]?

Engineering

1 answer:

Sliva [168]3 years ago

6 0

Answer:

11.34 g/cm3

Explanation:

At room temperature, where it is in a solid state, it is 11.34 $\frac{g}{cm^{3}}$ . While at melting temperature, at 327.5 ° C, it is 10.66 $\frac{g}{cm^{3}}$

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Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra

VashaNatasha [74]

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

int *p;

}arr[1000];

2) struct array{

int *p;

}arr[1000];

6 0

3 years ago

How does emotion affect a persons driving

Elodia [21]

Answer:

if ur mad you may drive faster if ur sad u may drive slower due to the amount of adrenaline and dopamine levels in your body in that given moment

Explanation:

4 0

3 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago

A hypereutectoid steel often presents hard and brittle cementite along the grain boundaries of pearlite. Which of the following

Anvisha [2.4K]

Answer:

(d) Spheroidizing

Explanation:

Spheroidizing

This is the heat treatment process for steel which having carbon percentage more than 0.8 %.As we know that a hard and brittle material is having carbon percentage more than 0.8 %.That is why this process is suitable for the hard materials.

In this process a hard and brittle materials convert into soft and ductile after this it improve the machine ability as well as improve the tool life.

In this process grain become spheroidal and these grains are ductile.

6 0

3 years ago

Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin

Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2 - $\sqrt{}$ ( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy = tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2 - $\sqrt{}$ ( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0

2 years ago