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zaharov [31]
3 years ago
10

We can model a certain battery as a voltage source in series with a resistance. The open-circuit voltage of the battery is 10 V

. When a 120-Ω resistor is placed across the terminals of the battery, the voltage drops to 7 V. Determine the internal resistance of the battery.
Engineering
1 answer:
Angelina_Jolie [31]3 years ago
3 0

Answer:

51.4 Ohms

Explanation:

By applying voltage division rule

V_f=v_i\times \frac {R_l}{R_l+R_m} where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then

7=10*\frac {120}{120+R_m}\\7R_m+840=1200\\R_m={1200-840}{7}=51.428571\approx 51.4 Ohms

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Paint can shaker mechanisms are common in paint and hardware stores. While they do a good job of mixing the paint, they are also
Ymorist [56]

Answer:

A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:

The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.

Explanation:

First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.

6 0
3 years ago
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
sergiy2304 [10]

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

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

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(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,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(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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Answer:

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ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m

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