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3 years ago

NASA SPACE SHUTTLE QUESTION:

Engineering

1 answer:

vfiekz [6]3 years ago

8 0

Explanation:

The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.

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1. What is resilience engineering and how does it pertain to the current<br>corona virus crisis?

saw5 [17]

Answer:

the ability to absorb or avoid damage without suffering complete failure and is an objective of design, maintenance and restoration for buildings and infrastructure

Explanation:

8 0

3 years ago

Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should w

Setler79 [48]

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

$Re_{5}$ = $Re_{12}$

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$ = $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

$Re_{5}$ = Reynold number of water pipe

$Re_{12}$ = Reynold number of oil pipe

$V_{5}$ = Velocity of water 5 diameter pipe = ?

$V_{12}$ = Velocity of oil 12 diameter pipe = 2.30

$v_{5}$ = Kinetic Viscosity of water = 1 x $10^{-6}$ $m^{2}$ /s

$v_{12}$ = Kinetic Viscosity of oil = 4 x $10^{-6}$ $m^{2}$ /s

$D_{5}$ = Diameter of pipe used for water = 0.05 m

$D_{12}$ = Diameter of pipe used for oil = 0.12 m

Use the formula

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$ = $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

By Removing square rots on both sides

${\frac{V_{5}XD_{5} }{v_{5}}}$ = ${\frac{V_{12}XD_{12} }{v_{12}}}$

${V_{5}$ = ${\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}$ x $v_{5}$

${V_{5}$ = [ (0.23 x 0.12m ) / (4 x $10^{-6}$ $m^{2}$ /s) x 0.05 ] 1 x $10^{-6}$ $m^{2}$ /s

${V_{5}$ = 1.38 m/s

4 0

3 years ago

Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What

Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature $T_1$ = 15°C

Initial Pressure $P_1$ = 5 atm

Final Pressure $P_2$ = 1 atm

Data obtain from steam tables of saturated water at 15°C are as follows:

Specific volume v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

$\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$ $\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \ K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

$W_{lost} = T_{surr} *S$

where ;

$T_{surr}$ is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K

From above the change in entropy is:

$\delta \ S = C_p \ In (\frac{T+ \delta \ T }{T}) * \beta V \delta P$

$\delta \ S = 4.19* \ In (\frac{288.15+0.093 }{288.15}) - 1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})$

$\delta \ S =1.408*10^{-3} \ J/gm.K$

$W_{lost} = T_{surr} *S$

$W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K$

$W_{lost} = 0.413 \ kJ/kg$

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0

4 years ago

andreev551 [17]

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks = 512 blocks

(b) Tag is 18

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No . of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index +block offset

So , 32 =tag+9+5

tag = 32-(9+5)

So Tag is 18

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

= 223

=8388608 bytes

7 0

4 years ago

Write a statement that increases numPeople by 5. Ex: If numPeople is initially 10, the output is: There are 15 people.

BartSMP [9]

Answer:

import java.util.Scanner;

public class AssigningNumberToVariable {

public static void main (String [] args) {

int numPeople;

int numPeople_inc;

numPeople = 10;

/* solution */

numPeople_inc=numPeople+5;

System.out.print("There are ");

System.out.print(numPeople_inc);

System.out.println(" people.");

}

Explanation:

1. Declare a new variable called: numPeople_inc

2. Add 5 each time: numPeople_inc=numPeople+5;

3. Print the answer: numPeople_inc=numPeople+5;

Optional, just write the line in the place where says: " your solutions goes here":

numPeople = numPeople+5;

8 0

3 years ago