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mojhsa [17]
3 years ago
13

If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B

Engineering
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

105 km

Explanation:

The motorist was going 30 km/hr, and it took 3 hours 30 minutes. That's 3.5 hours. 3.5×30=105

You might be interested in
Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload
Vinil7 [7]

Answer:

1) V_o = 10 liters

2) V_o = 12.26 liters

Explanation:

For isothermal process n =1

V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}

V_o = 10 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.03

actual \ volume = c1\times 10 = 10.3 liters

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.15

actual \volume = c1\times 10 = 11.5 liters

8 0
3 years ago
An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to =  4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × 10^{6} + 4  × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × 10^{6} - 4  × 10³

lower side frequencies =  3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies  - lower side frequencies

total width bandwidth = 8kHz

6 0
3 years ago
Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,
iren [92.7K]

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA,  IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters          IE       IC           IB            VE       VB         Vc

Unit                     mA     mA          μA            V           V          V

Value                  4.65    4.6039   46.039    10.7      10     4.6039

4 0
3 years ago
java Your program class should be called RomanNumerals Write a program that asks the user to enter a number within the range of
Ann [662]

Answer:

// Scanner class is imported to allow program

// receive input

import java.util.Scanner;

// RomanNumerals class is defined

public class RomanNumerals {

   // main method that signify beginning of program execution

   public static void main(String args[]) {

       // Scanner object scan is created

       // it receive input via keyboard

       Scanner scan = new Scanner(System.in);

       // Prompt is display asking the user to enter number

       System.out.println("Enter your number: ");

       // the user input is stored at numberOfOrder

       int number = scan.nextInt();

     

           // switch statement which takes number as argument

           // the switch statement output the correct roman numeral

           // depending on user input

          switch(number){

           case 1:

               System.out.println("I");

               break;

           case 2:

               System.out.println("II");

               break;

           case 3:

               System.out.println("III");

               break;

           case 4:

               System.out.println("IV");

               break;

           case 5:

               System.out.println("V");

               break;

           case 6:

               System.out.println("VI");

               break;

           case 7:

               System.out.println("VII");

               break;

           case 8:

               System.out.println("VIII");

               break;

           case 9:

               System.out.println("IX");

               break;

           case 10:

               System.out.println("X");

               break;

           // this part is executed if user input is not between 1 to 10

           default:

               System.out.println("Error. Number must be between 1 - 10.");

     }

   }

}

Explanation:

The program is well commented. A sample image of program output is attached.

The switch statement takes the user input (number) as argument as it goes through each case block in the switch statement and match with the corresponding case to output the roman version of that number. If the number is greater 10 or less than 1; the default block is executed and it display an error message telling the user that number must be between 1 - 10.

6 0
3 years ago
Read 2 more answers
3. A competency-based training program is
Vesna [10]
C and B is the answer of the questions
7 0
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