What can you add to a seatbelt ??<br> HELP ASAP

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

2 years ago

A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota

Soloha48 [4]

Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius. The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

m=((0.015m*0.200m)*0.6m)*(7800 kg/m^3)=28.08 kg

The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa= 10000000Pa+397887Pa= 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d =2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

5 0

3 years ago

Using a queue.

beks73 [17]

Answer:

// Radix Sort

#include<iostream>

using namespace std;

// function to get max value

int getMaximum(int array[], int n)

{

int max = array[0];

for (int i = 1; i < n; i++)

if (array[i] > max)

max = array[i];

return max;

}

// function to get min value

int getMinimum(int array[], int n)

{

int mn = array[0];

for (int i = 1; i < n; i++)

if (array[i] <mn)

mn= array[i];

return mn;

}

//counting sort

void counterSort(int array[], int n, int expo)

{

int out[100]; // out max

int i, count[10] = {0};

// Store count of occurrences in count[]

for (i = 0; i < n; i++)

count[ (array[i]/expo)%10 ]++;

// Change count[i] so that count[i] now contains actual

// position of this digit in out[]

for (i = 1; i < 10; i++)

count[i] += count[i - 1];

// construct the out max

for (i = n - 1; i >= 0; i--)

{

out[count[ (array[i]/expo)%10 ] - 1] = array[i];

count[ (array[i]/expo)%10 ]--;

}

for (i = 0; i < n; i++)

array[i] = out[i];

}

void counterSortDesc(int array[], int n, int expo)

{

int out[100]; // out max

int i, count[10] = {0};

// Store count of occurrences in count[]

for (i = 0; i < n; i++)

count[ (array[i]/expo)%10 ]++;

for (i = 10; i >=1; i--)

count[i] += count[i - 1];

// construct out max

for (i = 0; i >= n-1; i++)

{

out[count[ (array[i]/expo)%10 ] - 1] = array[i];

count[ (array[i]/expo)%10 ]++;

}

// Copy the out max to array[], so that array[] now

// contains sorted numbers according to current digit

for (i = 0; i < n; i++)

array[i] = out[i];

}

// Radix Sort function

void radixsort(int array[], int n)

{

// get maximum number

int m = getMaximum(array, n);

for (int expo = 1; m/expo > 0; expo *= 10)

counterSort(array, n, expo);

}

void radixsortDesc(int array[], int n)

{

// get minimum number

int m = getMinimum(array, n);

for (int expo = 1; m/expo > 0; expo *= 10)

counterSortDesc(array, n, expo);

}

// print an max

void print(int array[], int n)

{ cout<<"\n";

for (int i = 0; i < n; i++)

cout << array[i] << " ";

}

// Main function

int main()

{

int array[] = {185, 25, 35, 90, 904, 34, 2, 66};

int n = sizeof(array)/sizeof(array[0]);

radixsort(array , n);

print(array, n);

radixsortDesc(array,n);

print(array,n);

return 0;

}

Explanation:

7 0

2 years ago

Container need to be inspected

Sedaia [141]

Answer:

than look inside it

Explanation:

well if you need to inspect something, looking is very important

7 0

2 years ago

PLZ HELP ME!! Can someone plz give me a summary on engineering in 3-5 sentences.

myrzilka [38]

Engineering is about building things, learning how to build things and it being fun

4 0

3 years ago

What can you add to a seatbelt ?? HELP ASAP