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3 years ago

Static electricity is the result of

Engineering

1 answer:

grin007 [14]3 years ago

8 0

Answer:

Explanation:

Static electricity is the result of an imbalance between negative and positive charges in an object. These charges can build up on the surface of an object until they find a way to be released or discharged.

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4. Which type of duct undergoes more rigorous testing before it's labeled?

MakcuM [25]

Answer: Flexible duct because, They also developed more than 1,000 safety criteria, some of which are... labelled goods bearing a silver UL sticker on the duct that defines this listing. ... lie in the research carried out by UL and, as a result, the authorized use of the code. Air Ducts are required to perform 15 separate tests.

Source: https://dundasjafine.com/ul-classification/

3 0

3 years ago

Find the compressibility factor Z for oxygen at 3 MPa and 160 K.

saveliy_v [14]

Answer:

Z= 0.868

Explanation:

Given that

P= 3 MPa

T = 160 K

We know that

P v= Z R T

P= Pressure

v = specific volume

R= gas constant

T = Absolute temperature

Z= Compressibility factor

Here specific volume of gas is not given so we assume that specific volume gas

$v=0.012\ m^3/kg$

We know that for oxygen gas constant

R = 0.259 KJ/kg.K

Now by putting the values

P v = Z R T

3000 x 0.012 = Z x 0.259 x 160

Z= 0.868

So Compressibility factor is 0.868.

5 0

3 years ago

Write a function which multiplies the values in odd position values by 10. Odd positions in this case refers to the first value

xxMikexx [17]

Answer:

Using linkedlist on C++, we have the program below.

Explanation:

#include<iostream>

#include<cstdlib>

using namespace std;

//structure of linked list

struct linkedList

{

int data;

struct linkedList *next;

};

//print linked list

void printList(struct linkedList *head)

{

linkedList *t=head;

while(t!=NULL)

{

cout<<t->data;

if(t->next!=NULL)

cout<<" -> ";

t=t->next;

}

//insert newnode at head of linked List

struct linkedList* insert(struct linkedList *head,int data)

{

linkedList *newnode=new linkedList;

newnode->data=data;

newnode->next=NULL;

if(head==NULL)

head=newnode;

else

{

struct linkedList *temp=head;

while(temp->next!=NULL)

temp=temp->next;

temp->next=newnode;

}

return head;

}

void multiplyOddPosition(struct linkedList *head)

{

struct linkedList *temp=head;

while(temp!=NULL)

{

temp->data = temp->data*10; //multiply values at odd position by 10

temp = temp->next;

//skip odd position values

if(temp!= NULL)

temp = temp->next;

}

int main()

{

int n,data;

linkedList *head=NULL;

// create linked list

head=insert(head,20);

head=insert(head,5);

head=insert(head,11);

head=insert(head,17);

head=insert(head,23);

head=insert(head,12);

head=insert(head,4);

head=insert(head,21);

cout<<"\nLinked List : ";

printList(head); //print list

multiplyOddPosition(head);

cout<<"\nLinked List After Multiply by 10 at odd position : ";

printList(head); //print list

return 0;

}

5 0

3 years ago

The solid spindle AB is connected to the hollow sleeve CD by a rigid plate at C. The spindle is composed of steel (Gs = 11.2 x 1

dalvyx [7]

Answer:

T_max = 12.63 kip.in

Ф_a = 1.093°

Explanation:

Given:

- The modulus of rigidity of solid spindle G_ab = 11.2 * 10^6 psi

- The diameter of solid spindle d_ab = 1.75 in

- The allowable stress in solid spindle τ_ab = 12 ksi

- The modulus of rigidity of sleeve G_cd = 5.6 * 10^6 psi

- The outer diameter of sleeve d_cd = 3 in

- The thickness of sleeve t = 0.25

- The allowable stress in sleeve τ_cd = 7 ksi

Find:

- The largest torque T that can be applied to end A that does not exceed allowable stresses and sleeve angle of twist 0.375°

- The corresponding angle through which end A rotates.

Solution:

- Calculate the polar moment of inertia of both spindle AB and sleeve CD.

Spindle AB: c_ab = 0.5*d_ab = 0.5(1.75) = 0.875 in

J_ab = pi/2 c^4 = pi/2 0.875^4 = 0.92077 in^4

Sleeve CD: c_cd1 = 0.5*d_cd = 0.5(3) = 1.5 in , c_cd2 = c_cd1 - t = 1.25

J_cd = pi/2 (c_cd1^4 - c_cd2^4)= pi/2(1.5^4-1.25^4) = 4.1172 in^4

- The stress criteria for maximum allowable torque in spindle AB:

T_ab = J_ab*τ_ab / c_ab

T_ab = 0.92077*12 / 0.875

T_ab = 12.63 kip.in

- The stress criteria for maximum allowable torque in sleeve CD:

T_cd = J_cd*τ_cd / c_cd1

T_cd = 4.1172*7 / 1.5

T_cd = 19.21 kip.in

- The angle of twist criteria for point D:

T_d = J_cd*G_cd*Ф / L

T_d = 4.1172*5.6*10^6*0.006545 / 8

T_d = 18.86 kip.in

- The maximum allowable Torque for the structure is:

T_max = min ( 12.63 , 19.21 , 18.86 )

T_max = 12.63 kip.in

- The angle of twist of end A:

Ф_a = Ф_a/d = Ф_a/b + Ф_c/d:

T_max* ( L_ab / J_ab*G_ab + L_cd / J_cd*G_cd)

12.63*(12/0.92*11.2*10^6 + 8/4.117*5.6*10^6)

0.01908 rads = 1.093°

3 0

3 years ago

In a vapor-compression refrigeration cycle, ammonia exits the evaporator as saturated vapor at -10°C. The refrigerant enters the

sergij07 [2.7K]

Answer: (a) 0,142 (b) 52.99 (c) 2.83 (d) 88.26

Explanation:

If the refrigarating capacity is 150kw

(a) the mass flow rate of refrigerant, in kilograms per second is 0.142

(b) the power input to the compressor, in kilowatts is 52.99

(d) the isentropic compressor efficiency is 68.6 per cent

8 0

3 years ago