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3 years ago

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A teenager was pulling a prank and placed a large stuffed penguin in the middle of a roadway. A driver is traveling on this leve

l roadway and sees the penguin 525 feet ahead.

1 answer:

Anvisha [2.4K]3 years ago

3 0

Whats the question????

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How many kg moles of Sodium Sulphate will contain 10 kg of<br> Sodium?

enyata [817]

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(as per my thinking)

5 0

2 years ago

Read 2 more answers

Write a modular program that finds the equation, area, and circumference of a circle, given the coordinates of the center of the

Semenov [28]

Answer:

Explanation:

#include <bits/stdc++.h>

#include <iostream>

#include <string>

#include <cmath>

using namespace std;

//this function reads the cooridnates of Center from the user

//parameteres are pointer variables of Cx,Cy

//it does not return anything and stores coordinates at given addresses of Cx,Cy

void readCenter(int *Cx,int *Cy)

{

string cooridnates;

cout << "Enter the x and y cooridnates of the centre of the circle separated by comma: " ;

getline(cin,cooridnates);//reading inputs

//convering string ot integer

string x = cooridnates.substr(0, cooridnates.find(","));

string y = cooridnates.substr(cooridnates.find(",")+1);

*Cx=stoi(x);

*Cy=stoi(y);

}

//this function reads the cooridnates of Point from the user

//parameteres are pointer variables of Px,Py

//it does not return anything and stores coordinates at given addresses of Px,Py

void readPoint(int *Px,int *Py)

{

string cooridnates;

cout << "Enter the x and y cooridnates of a point on the circle separated by comma: " ;

getline(cin,cooridnates);//reading inputs

//convering string ot integer

string x = cooridnates.substr(0, cooridnates.find(","));

string y = cooridnates.substr(cooridnates.find(",")+1);

*Px=stoi(x);

*Py=stoi(y);

}

double findArea(double radius)

{

double pi=acos(-1);

return pi*pow(radius,2);

}

double findCircum(double radius)

{

double pi=acos(-1);

return 2*pi*radius;

}

int main()

{

int Cx,Cy;

int Px,Py;

readCenter(&Px,&Py);

readPoint(&Cx,&Cy);

double radius=sqrt(pow((Px-Cx),2)+pow((Py-Cy),2));

cout<<"The circle has an area of "<<findArea(radius)<<" sqaure units\n";

cout<<"The circle has a Circumference of "<<findCircum(radius)<<" units";

return 0;

}

3 0

3 years ago

A master stud pattern is laid out somewhat<br> like a?

Svetradugi [14.3K]

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

3 0

1 year ago

). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially

Delicious77 [7]

Answer:

$\frac{e'_z}{e_z} = 0.87142$

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

εx = - [ σx - ν( σy + σz ) ] / E

εy = - [ σy - ν( σx + σz ) ] / E

εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

εx = [ ν*σz ] / E

εy = [ ν*σz ] / E

εz = - [ σz ] / E .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E

εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note: σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2

σx' ( 1 - v ) = [ ν*σz ] / 2

σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

εz' = - [ σz - ν( σy' + σx' ) ] / E

εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

$\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142$ ... Answer

5 0

2 years ago

Witch measuring tool would be used to determine the diameter of a crankshaft journal

djverab [1.8K]

Answer =

dial bore gauge

a “dial bore gauge” measures the inside of round holes, such as the bearing journals . can mesure up to 2” and 6” diameter holes .

when ( “ ) is next to a number it means inches fwi - but hope this helped have a good day :)

3 0

2 years ago