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Harrizon [31]
2 years ago
8

How many times greater is the value of the 2 of the 270413 than the valuce of the 2 in 419427?

Engineering
1 answer:
nadezda [96]2 years ago
7 0
The value of 2 in 270,413 is 10,000x greater than the value of 2 in 419,427.
Step-by-step explanation:
The value of 2 in 270,413 is 200,000.
The value of 2 in 419,427 is 20
Divide the two numbers together to find your answer:
200,000/20 = 10,000
The value of 2 in 270,413 is 10,000x greater than the value of 2 in 419,427.

Please mark it as brainlest answer:).
You might be interested in
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:

8 0
3 years ago
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

5 0
3 years ago
Explain the difference between the connection of a cumulative compound and a differential compound motor
Alexandra [31]

Answer:

Explanation:

A motor is a device that directs current in electrical energy form to mechanical energy, which is known as direct current (DC) motors.

DC motors are of three types: (a) The series motor, (b) The shunt motor, and (c) the compound motor. Our main focus here is the Compound motor, which is further sub-divided into:

i) The cumulative compound motors

ii) The differential compound motors

The difference between these two are:

Cumulative compound motors                  Differential compound motors

In cumulative compound motors,              In differential compound motors,

both the series and shunt windings          both series and shunt are

are connected in a way that,                     connected in a way that the

production of fluxes through them           production of fluxes via them

assist each other i.e. they aid each          always opposes each other i.e.

other in the production of magnetism      they oppose each other in the

                                                                    production of magnetism.

6 0
3 years ago
Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and
Alborosie

Answer:

V_{pp}=2V

Explanation:

Source Voltage V= 120V

Frequency f=60Hz

Peak output voltage Vp=15V

Peak Output Voltage with filter V_p'=14V

Generally the equation for Peak to peak voltage is mathematically given by

V_p'=V_p-\frac{V_{pp}}{2}

Therefore

V_{pp}=2(V_p-v_p')

V_{pp}=2(15-14)

V_{pp}=2V

5 0
2 years ago
R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its
attashe74 [19]

Answer:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

<u>Calculate the change in the enthalpy of R-134a </u>

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant  

hence ; s1 = s2

by interpolation  ; h2 = 106.95

Δ enthalpy = h2 - h1

                  =  ( 106.95 - 129.85 ) = -23 Btu/Ibm

5 0
2 years ago
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