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ludmilkaskok [199]
3 years ago
14

(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific

weight of 66.4 lb/ft3. Determine the vertical force required to give the tank an upward acceleration of 10.7 ft/s2.
Engineering
1 answer:
lina2011 [118]3 years ago
5 0

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content =  66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

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Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA
otez555 [7]

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

  q=(struct node *)malloc(sizeof(struct node)*count);

  dfa->initialStateID = -1;

  dfa->presentStateID = -1;

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  return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

  int tID;

  for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

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      {

          dfa->presentStateID = pPresentState->transitions[tID].toStateID;

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      }

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  dfa->presentStateID = pPresentState->defaultToStateID;

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void State_add (DFA* pDFA, DFAState* newState)

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  newState->ID = pDFA->numOfStates;

  pDFA->states[pDFA->numOfStates] = newState;

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}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

  DFAState* state = dfa->states[fromStateID];

  state->transitions[state->numOfTransitions].toStateID = toStateID;

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void reset(DFA* dfa)

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5 0
3 years ago
A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro
Zarrin [17]

Answer:

  83.6%

Explanation:

<h3>(a)</h3>

On its current schedule, the plant can theoretically produce ...

  (30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

<h3>(b)</h3>

On the proposed schedule, this production becomes ...

  (30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

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The increase in capacity is ...

  (2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

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