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velikii [3]
3 years ago
11

What is 100 kPa in psia?

Engineering
1 answer:
Korolek [52]3 years ago
8 0

Answer:

The pressure in pounds per square inch (psi) equals 14.50 psi.

Explanation:

Since psi stands for 'pounds per square inch' and Pascals stands for 'Newtons per square meters'

We can write the given pressure as

Pressure=100kPa\\\\Pressure=100\times 10^{3}N/m^{2}

Now since 1 Newton of force equals 0.22481 Pound force also

since 1 meter equals 39.37 inches

Using the above values we get

Pressure=100\times 10^{3}\times \frac{0.22481lb}{(39.37)^{2}inch^{2}}

thus the pressure becomes

Pressure=100\times 10^{3}\times \frac{0.22481}{(39.31)^{2}}pounds/inch^{2}\\\\Pressure=14.50psi

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/* Function findBestVacation * duration: number of vacation days * prefs: prefs[k] indicates the rate specified for game k * pla
alexira [117]

Answer:

This is the C++ code for the above problem:

#include<bits/stdc++.h>

using namespace std;

int computeFunLevel(int start, int duration, int prefs[], int ngames, int plan[]) {

if(start + duration > 365) { //this is to check wether duration is more than total no. of vaccation days

return -1;

}

int funLevel = 0;

for(int i=start; i<start+duration; i++) { //this loop runs from starting point till

//start + duration to sum all the funlevel in plan.

funLevel = funLevel + prefs[plan[i]];

}

return funLevel;

}

int findBestVacation(int duration, int prefs[], int ngames, int plan[]) {

int max = 0, index = 0, sum = 0 ;

for(int i=1; i<11; i++){ //this loop is to run through whole plan arry

sum = 0; //sum is initialized with zero for every call in plan ,

//in this case loop should run to 366,but for example it is 11

//as my size of plan array is 11

for(int j=0; j<duration; j++) { // this loop is for that index to index+duration to calc

//fun from that index

sum = sum + prefs[plan[i]];

}

if(sum>max) { //this is to check max funlevel and update the index from which max fun can be achieved

max = sum;

index = i;

}

}

return index;

}

int main() {

int ngames = 5;

int prefs[] = { 1,2,0,5,2 };

int plan[] = { 0,2,0,3,3,4,0,1,2,3,3 };

int start = 1;

int duration = 4;

cout << computeFunLevel(start, duration, prefs, ngames, plan) << endl;

cout << computeFunLevel(start, 555, prefs, ngames, plan) << endl;

cout << findBestVacation(4, prefs, ngames, plan) << endl;

}

The screen of the program is given below.

3 0
2 years ago
Ronny wants to calculate the mechanical advantage. He needs to determine the length of the effort arm and the length of the load
kakasveta [241]

Answer:

I hope it's helpful.

Explanation:

Simple Machines

Experiments focus on addressing areas pertaining to the relationships between effort force, load force, work, and mechanical advantage, such as: how simple machines change the force needed to lift a load; mechanical advantages relation to effort and load forces; how the relationship between the fulcrum, effort and load affect the force needed to lift a load; how mechanical advantage relates to effort and load forces and the length of effort and load arms.

Through investigations and models created with pulleys and levers, students find that work in physical terms is a force applied over a distance. Students also discover that while a simple machine may make work seem easier, in reality the amount of work does not decrease. Instead, machines make work seem easier by changing the direction of a force or by providing mechanical advantage as a ratio of load force to effort force.

Students examine how pulleys can be used alone or in combination affect the amount of force needed to lift a load in a bucket. Students find that a single pulley does not improve mechanical advantage, yet makes the effort applied to the load seem less because the pulley allows the effort to be applied in the direction of the force of gravity rather than against it. Students also discover that using two pulleys provides a mechanical advantage of 2, but that the effort must be applied over twice the distance in order to gain this mechanical advantage Thus the amount of work done on the load force remains the same.

Students conduct a series of experiments comparing the effects of changing load and effort force distances for the three classes of levers. Students discover that when the fulcrum is between the load and the effort (first class lever), moving the fulcrum closer to the load increases the length of the effort arm and decreases the length of the load arm. This change in fulcrum position results in an increase in mechanical advantage by decreasing the amount of effort force needed to lift the load. Thus, students will discover that mechanical advantage in levers can be determined either as the ratio of load force to effort force, or as the ratio of effort arm length to load arm length. Students then predict and test the effect of moving the fulcrum closer to the effort force. Students find that as the length of the effort arm decreases the amount of effort force required to lift the load increases.

Students explore how the position of the fulcrum and the length of the effort and load arms in a second-class lever affect mechanical advantage. A second-class lever is one in which the load is located between the fulcrum and the effort. In a second-class lever, moving the load changes the length of the load arm but has no effect on the length of the effort arm. As the effort arm is always longer than the load arm in this type of lever, mechanical advantage decreases as the length of the load arm approaches the length of the effort arm, yet will always be greater than 1 because the load must be located between the fulcrum and the effort.

Students then discover that the reverse is true when they create a third-class lever by placing the effort between the load and the fulcrum. Students discover that in the case of a third-class lever the effort arm is always shorter than the load arm, and thus the mechanical advantage will always be less than 1. Students also create a model of a third-class lever that is part of their daily life by modeling a human arm.

The CELL culminates with a performance assessment that asks students to apply their knowledge of simple machine design and mechanical advantage to create two machines, each with a mechanical advantage greater than 1.3. In doing so, students will demonstrate their understanding of the relationships between effort force, load force, pulleys, levers, mechanical advantage and work. The performance assessment will also provide students with an opportunity to hone their problem-solving skills as they test their knowledge.

Through this series of investigations students will come to understand that simple machines make work seem easier by changing the direction of an applied force as well as altering the mechanical advantage by afforded by using the machine.

Investigation focus:

Discover that simple machines make work seem easier by changing the force needed to lift a load.

Learn how effort and load forces affect the mechanical advantage of pulleys and levers.

8 0
2 years ago
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si
alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

<u>A) Classifying the soil according to USCS system</u>

 ( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0
3 years ago
Is the COP of a heat pump always larger than 1?
Liono4ka [1.6K]

Answer:

Yes

Explanation:

Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.

We know that

COP of heat pump=  1 + COP of refrigeration

It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.  

3 0
2 years ago
Because of ____________ people must make choices, and when they choose, they incur a(n)______________.
lbvjy [14]
There must be a photo for me to answer!
3 0
2 years ago
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