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Gennadij [26K]
3 years ago
14

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

neration potential of a wind turbine with 90-m-diameter (D) blades at that location. Take the air density to be 1.25 kg/m3.
Engineering
1 answer:
tangare [24]3 years ago
5 0

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

Diameter of the turbine = 90 m

Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

E=\dfrac{1}{2}mv^2\ J

That is why mechanical energy per unit mass will be

e=\dfrac{1}{2}v^2\ J/kg

Now by putting the values in the above equation we get

e=\dfrac{1}{2}\times 10^2\ J/kg

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

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. Chemical manufacturers must present which Information on the product's label?
STatiana [176]

Answer: A) Product identifier

Explanation:

A product identifier is a means, name or number used to identify hazardous and dangerous chemicals on a label or in the SDS( safety data sheet). It provides a special means which helps the user to identify the chemical. Any product identifier used should permit a cross-references to be made among the listed hazardous chemicals required in the written hazard communication program, and the label including SDS(safety data sheet).

5 0
3 years ago
import java.util.Scanner; public class FindSpecialValue { public static void main (String [] args) { Scanner scnr = new Scanner(
Hitman42 [59]

Answer:

Java program explained below

Explanation:

FindSpecialNumber.java

import java.util.Scanner;

public class FindSpecialNumber {

public static void main(String[] args) {

//Declaring variable

int number;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

//getting the input entered by the user

System.out.print("Enter a number :");

number = sc.nextInt();

/* Based on user entered number

* check whether it is special number or not

*/

if (number == -99 || number == 0 || number == 44) {

System.out.println("Special Number");

} else {

System.out.println("Not Special Number");

}

}

}

_______________

Output#1:

Enter a number :-99

Special Number

Output#2:

Enter a number :49

Not Special Number

7 0
2 years ago
Ok there........................................................................
Juliette [100K]

Answer:

ok THERE

Explanation:

4 0
3 years ago
Why is it important to know the accuracy and precision of a measuring device? Do you think that the dial caliper manufacturer’s
Leto [7]

Answer:

Accuracy and precision allow us to know how much we can rely on a measuring device readings. ±.001 as a "accuracy" claim is vague because there is no unit next to the figure and the claim fits better to the definition of precision.

Explanation:

Accuracy and Precision: the golden couple.

Accuracy and precision are key elements to define if a measuring device is reliable or not for a specific task. Accuracy determines how close are the readings from the ideal/calculated values. On the other hand, precision refers to repeatability, that is to say how constant the readings of a device are when measuring the same element at different times. One of those two key concepts may not fulfill the criteria for measuring tool to be used on certain engineering projects where lack of accuracy (disntant values from real ones) or precision (not constant readings) may lead to malfunctons and severe delays on the project development.

±.001 what unit?

The manufacturer says that is an accuracy indicator, nevertheless there is now unit stated so this is not useful to see how accurate the device is. Additionally, That notation is more used to refer to device tolerances, that is to say the range of possible values the instrument may show when reading and element. It means it tells us more about the device precision during measurments than actual accuracy. I would recommend the following to the dial calipers manufacturers to better explain its measurement specifications:

  1. Use  ±.001 as  a reference for precision. It is important to add the respective unit for that figure.
  2. Condcut test to define the actual accuracy value an present it using one of the common used units for that:  Error percentage or ppm.

3 0
3 years ago
An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a l
Mekhanik [1.2K]

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

6 0
3 years ago
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