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2 years ago

State and explain briefly four (4) functions that a machine foundation should fulfill.

Engineering

1 answer:

MrMuchimi2 years ago

8 0

Answer:

1 The weight of the foundation block should be enough to withstand vibrations

2 The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.

3.The base should be sufficiently rigid to have the requisite rigidity

4 a distance should be created all across machine foundation to separate it from the adjacent parts of the building

Explanation:

1. The weight of the foundation block should be enough to withstand vibrations and to avoid friction between device and the surrounding soil as well. This can be done by increasing the base block weight in supporting with engine power

2.The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.

3.The base should be sufficiently rigid to have the requisite rigidity as the slightest misdirection of foundation may cause significant bearing disorders.

4.To avoid propagation of vibration from a device to the adjacent parts of the building, a distance should be created all across machine foundation to separate it from the adjacent parts of the building.

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A person puts a few apples into the freezer at -15oC to cool them quickly for guests who are about to arrive. Initially, the app

frosja888 [35]

Answer:

Temperature at center of apples = 11.2⁰C

Temperature at surface of apples = 2.7⁰C

Amount of Heat transferred = 17.2kJ

Explanation:

The properties of apple are given as:

k = 0.418 W/m.°C

ρ = 840 kg/m³

Cр = 3.81 kJ/kg.°C

α = 1.3*10 ⁻⁷ m²/s

h = 8 W/m².°C

d = 0.09m

r = 0.045m

t = 1 hour = 3600s

<h2>Solution</h2>

Biot number is given as:

$Bi = \frac{hr}{k}= \frac{8\cdot0.045}{0.418}=0.861$

The constants λ₁ and A₁ corresponding to Biot number (from the table) are:

λ₁ = 1.476

A₁ = 1.239

Fourier Number is:

$T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2$

As Fourier Number > 0.2 , one term approximates solutions are applicable

The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.

8 0

2 years ago

Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made

Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random.Therefore, number ways in which cars can line up is given as:

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

<u>Probability = 0.00001 = 0.001%</u>

4 0

2 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

2 years ago

___________ is NOT a common injury that an automotive tech may experience at work.

Degger [83]

Answer:The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

Explanation:

The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

7 0

2 years ago

Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user

Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

final int NUM_VALS = 4;

int[] userValues = new int[NUM_VALS];

int i;

int matchValue;

int numMatches = -99; // Assign numMatches with 0 before your for loop

matchValue = scnr.nextInt();

for (i = 0; i < userValues.length; ++i) {

userValues[i] = scnr.nextInt();

}

/* Your solution goes here */

numMatches = 0;

for (i = 0; i < userValues.length; ++i) {

if(userValues[i] == matchValue) {

numMatches++;

}

System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

}

8 0

2 years ago