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4 years ago

1. Assume that EAX contains ff ff ff 51 and the doubleword referenced by value contains ff ff ff f1.

Engineering

1 answer:

IRINA_888 [86]4 years ago

6 0

Answer:

1. True

2. False

Explanation:

given data

EAX contains = ff ff ff 51

doubleword referenced = ff ff ff f1

conditional jump add = eax

solution

1st statement is true

but 2nd statement is false

as here

js or jne instruction is the conditional jump that is follow a test
It jump to the specified location when previous instructions are set the SF (Sign Flag) .

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A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

mash [69]

<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

$\\$

Explanation:

<h2>Given:</h2>

$a_1 \:=\:0.4\:m/s²$

$t_1 \:=\:60\:s$

$v_{i1} \:=\:0\:m/s$

$a_2 \:=\:0\:m/s²$

$t_2 \:=\:25\:min\:=\:1500\:s$

$a_3 \:=\:-0.8\:m/s²$

$v_{f3} \:=\:0\:m/s$

$\\$

<h2>Required:</h2>

Distance from Station A to Station B

$\\$

<h2>Equation:</h2>

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v\:=\:\frac{d}{t}$

$\\$

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for $v_{f1}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}$

$24\:m/s\:=\:v_f\:-\:0\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave1}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_1$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{60\:s}$

$720\:m\:=\:d$

$d_1\:=\:720\:m$

$\\$

<h3>Distance when a = 0 m/s²</h3>

$v_{f1}\:=\:v_{i2}$

$v_{i2}\:=\:24\:m/s$

$\\$

Solve for $v_{f2}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}$

$0\:=\:v_f\:-\:24\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave2}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:24\:m/s$

$\\$

Solve for $d_2$

$v\:=\:\frac{d}{t}$

$24\:m/s\:=\:\frac{d}{1500\:s}$

$36,000\:m\:=\:d$

$d_2\:=\:36,000\:m$

$\\$

<h3>Distance when a = -0.8 m/s²</h3>

$v_{f2}\:=\:v_{i3}$

$v_{i3}\:=\:24\:m/s$

$\\$

Solve for $v_{f3}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}$

$(t)(-0.8\:m/s²)\:=\:-24\:m/s$

$t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}$

$t\:=\:30\:s$

$\\$

Solve for $v_{ave3}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_3$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{30\:s}$

$360\:m\:=\:d$

$d_3\:=\:360\:m$

$\\$

<h3>Total Distance from Station A to Station B</h3>

$d\:= \:d_1\:+\:d_2\:+\:d_3$

$d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m$

$d\:= \:37,080\:m$

$d\:= \:37.08\:km$

$\\$

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

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3 years ago

Select three types of lines that engineers use to help represent the shape of a design in a sketch.

Vikki [24]

Hidden lines

Used to describe the in shown lines (like diagonals inside cubes)

Extension lines:-

Used to explain the expansion of structures like building

Object lines

Used to describe the structure of objects and the lining to show borders

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3 years ago

The following sentence can be categorized as what stage in the Scientific Method: Abraham notices that Saharan desert ants will

Marat540 [252]

Answer:

Saharan desert ants is making observations

Explanation:

The Saharan desert ants are using every available opportunity to make observations in search of food and that's the reason why they take great distances. However, after getting food, they are always eager to return back to their niche as soon as possible.

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4 years ago

Why is resume not consistent?

anzhelika [568]

Hehehehehehehehjsjdhsjakdkhdhahajdjdhfhsjjskskfkfjdjdififjsja sishfhsjdj

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3 years ago

The elementary liquid-phase series reaction

liraira [26]

Answer:

Concentration of A: $\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Concentration of B: $\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t } -e^{-k_{2}t } )$

Concentration of C: $\frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1} t}$

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

$r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}$

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

$-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Where x=CAo and y=CA

The mass balance for B:

$-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t }-ex^{-k_{2}t } )$

The mass balance for C:

$\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1}t }$

The maximum concentration of C is:

$C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1} }} =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}$

and the maximum time is:

$t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1} } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h$

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3 years ago