if a current of 5 amps flows through a resistance of 40 ohms, what is the voltage across that resistor

Saturated liquid water flows through 2 cm ID stainless steel tubes at 200 g/s. The water is at 80oC and the inside surface of th

EleoNora [17]

Answer:

steel

Explanation:

8 0

3 years ago

A six-lane freeway (three lanes in each direction) with a 5.5% uphill grade 1.5 miles long has 10-ft lanes and obstructions 5 ft

sertanlavr [38]

The answer would be:
A. LOS B

4 0

3 years ago

Read 2 more answers

C#: Arrays - Ask the user how many students names they want to store. You will create two parallel arrays (e.g. 2 arrays with th

zhenek [66]

Answer:

using System;
public class Program
{
public static void Main()
{
Console.WriteLine("Enter number of students: ");
int num = Convert.ToInt32(Console.ReadLine());
string [] firstName = new string[num];
string [] lastName = new string[num];
for(int i=0 ; i < num; i++){
Console.WriteLine("Enter first name: ");
firstName[i] = Console.ReadLine();
Console.WriteLine("Enter last name: ");
lastName[i] = Console.ReadLine();
}
for(int j=0; j < num; j++){
Console.WriteLine(lastName[j] + "," + firstName[j]);
}
}
}

Explanation:

Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).

Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).

Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).

4 0

3 years ago

Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0

bija089 [108]

Answer:

(a) 20 MHz

(b) 1.025 KW

(c) 3.33 ns

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = 20 MHz

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = 1.025 KW

(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = 3.33 ns

(d) 33 x10^(-12)F = 33 pico F = 33 pF

8 0

4 years ago

An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve

lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa

The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0

3 years ago