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Marizza181 [45]
3 years ago
10

At what voltage would a () capacitor have the minimum energy to raise by ignoring all losses in the system? If needed, you may a

ssume a gravitational acceleration of 9.80 m/s2.

Engineering
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

V = 280.15 V

Explanation:

" The complete question is attached with figure"

Given:

- The capacitance of the capacitor C = 10 nF

- The amount of mass attached to motor m = 4 grams

- The amount of distance it is to be lifted h = 1 cm

- Ignore all other losses in the system

Find:

- The voltage required to lift the mass m through distance h?

Solution:

- The conservation of energy for the entire system is written as:

                            Work_gravity = U_c

Where,

           Work_gravity: Work done by gravity on mass m

           U_c: The amount of energy stored in a capacitor

                             m*g*h = 0.5*C*V^2

                             V^2 = 2*m*g*h / C

                             V = sqrt ( 2*m*g*h / C )

Plug in the values:

                             V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )

                             V = sqrt ( 78,480)

                             V = 28.15 V

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A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp
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Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

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For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our w
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Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

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(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

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(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

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

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

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

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

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