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Vlada [557]
3 years ago
14

When a switch is closed, completing an lr series circuit, the time needed for the current to reach one half its maximum value is

_____ time constants?
Computers and Technology
1 answer:
uysha [10]3 years ago
7 0
The answer is <span>0.693 .  </span><span>When a switch is closed, completing an LR series circuit, the time needed for the current to reach one half its maximum value is 0.693   

</span>e^(-t/T) = 0.5 
<span>-t/T = ln(0.5) = -0.693 </span>
You might be interested in
Construct a class that will model a quadratic expression (ax^2 + bx + c). In addition to a constructor creating a quadratic expr
stich3 [128]

Answer:

Following are the code to this question:

#include <iostream>//header file

#include<math.h>//header file

using namespace std;

class Quadratic//defining a class Quadratic  

{

private:

double a,b,c;//defining a double variable

public:

Quadratic()//defining default constructor

{

a = 0;//assigning value 0  

b = 0;//assigning value 0  

c = 0;//assigning value 0  

}

Quadratic(double a, double b, double c)//defining a parameterized constructor  

{

this->a = a;//use this keyword to hold value in a variable

this->b = b;//use this keyword to hold value in b variable

this->c = c;//use this keyword to hold value in c variable

}

double getA() //defining a get method  

{

return a;//return value a

}

void setA(double a)//defining a set method to hold value in parameter

{

this->a = a;//assigning value in a variable

}

double getB() //defining a get method  

{

return b;//return value b

}

void setB(double b)//defining a set method to hold value in parameter  

{

this->b = b;//assigning value in b variable

}

double getC() //defining a get method

{

return c;//return value c

}

void setC(double c)//defining a set method to hold value in parameter

{

this->c = c;//assigning value in c variable

}

double Evaluate(double x)//defining a method Evaluate to hold value in parameter

{

return ((a*x*x)+(b*x)+c);//return evaluated value

}

double numberOfReal()//defining a method numberOfReal to calculates the real roots

{

return (b*b)-(4*a*c);//return real roots

}

void findroots()//defining a method findroots

{

double d=numberOfReal();//defining double variable to hold numberOfReal method value

if(d<0)//use if block to check value of d less than 0

cout<<"Equation has no real roots"<<endl;//print message

else

{

double r1=(-b+sqrt(numberOfReal()))/(2*a);//holding root value r1

double r2=(-b-sqrt(numberOfReal()))/(2*a);//holding root value r2

if(r1==r2)//defining if block to check r1 equal to r2

cout<<"Equation has one real root that is "<<r1<<endl;//print message with value

else//else block

cout<<"The equation has two real roots that are "<<r1<<" and "<<r2<<endl;////print message with value

}

}

void print()//defining a method print  

{

cout<< a << "x^2 + " << b << "x + " << c <<endl;//print Quadratic equation

}

};

int main()//defining main method  

{

Quadratic q(5,6,1);//creating Quadratic class object that calls parameterized constructor

q.print();//calling print method

cout<<q.numberOfReal()<<endl;//calling method numberOfReal that prints its value

q.findroots();//calling method findroots

cout<<q.Evaluate(-1);//calling method Evaluate that prints its value

return 0;

}

Output:

5x^2 + 6x + 1

16

The equation has two real roots that are -0.2 and -1

0

Explanation:

In the above code, a class "Quadratic" is declared, which is used to define a default and parameter constructor to holds its parameter value.

In the next step, the get and set method is defined that holds and returns the quadratic value, and "Evaluate, numberOfReal, findroots, and print" in the evaluate method a double variable is used as a parameter that returns evaluated value.

In the "numberOfReal" method it calculates the real roots and returns its value. In the "findroots" method a double variable "d" is declared that hold "numberOfReal" value,

and use a conditional statement to check its value, and in the print method, it prints the quadratic equation.

In the main method, the lass object it calls the parameterized constructor and other methods.

5 0
2 years ago
A camera on a tripod shows the ground floor of a skyscraper and then moves upward to show its entire length. what camera movemen
scZoUnD [109]

Answer:

A pedestal. It involves moving the camera upwards or downwards in relation to a subject.

8 0
2 years ago
3.What are the pros and cons of using a linked implementation of a sparse matrix, as opposed to an array-based implementation
weqwewe [10]

Answer:

speed and storage

Explanation:

The pros and cons of both are mainly in regards to speed and storage. Due to linked lists elements being connected to one another it requires that each previous element be accessed in order to arrive at a specific element. This makes it much slower than an array-based implementation where any element can be quickly accessed easily due to it having a specific location. This brings us to the other aspect which is memory. Since Arrays are saved as a single block, where each element has a specific location this takes much more space in the RAM as opposed to a linked implementation which is stored randomly and as indexes of the array itself.

Array Implementation:

  • Pros: Much Faster and Easier to target a specific element
  • Cons: Much More Space needed

Linked Implementation

  • Pros: Less overall space needed
  • Cons: Much slower speed.
5 0
3 years ago
Int, char, bool, and double are all valid data types in c , true or false
arsen [322]
False. bool is in C++, ints are used in C.

6 0
3 years ago
I need the SQL statements for these questions:
zimovet [89]

Answer:

Explanation:

/* From the information provided, For now will consider the name of table as TRIPGUIDES*/

/*In all the answers below, the syntax is based on Oracle SQL. In case of usage of other database queries, answer may vary to some extent*/

1.

Select R.Reservation_ID, R.Trip_ID , C.Customer_Num,C.Last_Name from Reservation R, Customer C where C.Customer_Num=R.Customer_Num ORDER BY C.Last_Name

/*idea is to select the join the two tables by comparing customer_id field in two tables as it is the only field which is common and then print the desired result later ordering by last name to get the results in sorted order*/

2.

Select R.Reservation_ID, R.Trip_ID , R.NUM_PERSONS from Reservation R, Customer C where C.Customer_Num=R.Customer_Num and C.LAST_NAME='Goff' and C.FIRST_NAME='Ryan'

/*Here, the explaination will be similar to the first query. Choose the desired columns from the tables, and join the two tables by equating the common field

*/

3.

Select T.TRIP_NAME from TRIP T,GUIDE G,TRIPGUIDES TG where T.TRIP_ID=TG.TRIP_ID and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Abrams' and G.FIRST_NAME='Miles'

/*

Here,we choose three tables TRIP,GUIDE and TRIPGUIDES. Here we selected those trips where we have guides as Miles Abrms in the GUIDES table and equated Trip_id from TRIPGUIDES to TRIP.TRIP_Name so that can have the desired results

*/

4.

Select T.TRIP_NAME

from TRIP T,TRIPGUIDES TG ,G.GUIDE

where T.TRIP_ID=TG.TRIP_ID and T.TYPE='Biking' and TG.GUIDE_NUM=G.GUIDE_NUM and G.LAST_NAME='Boyers' and G.FIRST_NAME='Rita'

/*

In the above question, we first selected the trip name from trip table. To put the condition we first make sure that all the three tables are connected properly. In order to do so, we have equated Guide_nums in guide and tripguides. and also equated trip_id in tripguides and trip. Then we equated names from guide tables and type from trip table for the desired results.

*/

5.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='2016-07-23' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

/*

The explaination for this one will be equivalent to the previous question where we just equated the desired columns where we equiated the desired columns in respective fields and also equated the common entities like trip ids and customer ids so that can join tables properly

*/

/*The comparison of dates in SQL depends on the format in which they are stored. In the upper case if the

dates are stored in the format as YYYY-MM-DD, then the above query mentioned will work. In case dates are stored in the form of a string then the following query will work.

SELECT C.LAST_NAME , T.TRIP_NAME , T.START_LOCATION FROM CUSTOMER C, TRIP T, RESERVATION R WHERE R.TRIP_DATE='7/23/2016' AND T.TRIP_ID=R.TRIP_ID AND C.CUSTOMER_NUM=R.CUSTOMER_NUM

*/

6.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION R WHERE R.TRIP_ID IN

{SELECT TRIP_ID FROM TRIP T WHERE STATE='ME'}

/*

In the above question, we firstly extracted all the trip id's which are having locations as maine. Now we have the list of all the trip_id's that have location maine. Now we just need to extract the reservation ids for the same which can be trivally done by simply using the in clause stating print all the tuples whose id's are there in the list of inner query. Remember, IN always checks in the set of values.

*/

7.

Select R.RESERVATION_ID, R.TRIP_ID,R.TRIP_DATE FROM RESERVATION WHERE

EXISTS {SELECT TRIP_ID FROM TRIP T WHERE STATE='ME' and R.TRIP_ID=T.TRIP_ID}

/*

Unlike IN, Exist returns either true or false based on existance of any tuple in the condition provided. In the question above, firstly we checked for the possibilities if there is a trip in state ME and TRIP_IDs are common. Then we selected reservation ID, trip ID and Trip dates for all queries that returns true for inner query

*/

8.

SELECT G.LAST_NAME,G.FIRST_NAME FROM GUIDE WHERE G.GUIDE_NUM IN

{

SELECT DISTINCT TG.GUIDE_NUM FROM TRIPGUIDES TG WHERE TG.TRIPID IN {

SELECT T.TRIP_ID FROM TRIP T WHERE T.TYPE='Paddling'

}

}

/*

We have used here double nested IN queries. Firstly we selected all the trips which had paddling type (from the inner most queries). Using the same, we get the list of guides,(basically got the list of guide_numbers) of all the guides eds which were on trips with trip id we got from the inner most queries. Now that we have all the guide_Nums that were on trip with type paddling, we can simply use the query select last name and first name of all the guides which are having guide nums in the list returned by middle query.

*/

4 0
2 years ago
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