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2 years ago

What Happens If A Sonic Boom Is Created?

Engineering

1 answer:

Volgvan2 years ago

8 0

Answer:

Explanation:

A sonic boom is a loud sound kind of like an explosion. It's caused by shock waves created by any object that travels through the air faster than the speed of sound. Sonic booms create huge amounts of sound energy. When an object moves through the air, it makes pressure waves in front of and behind it.

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The town of Mustang, TX is concerned that waste heat discharged from a new up- stream power plant will decimate the minnow popul

vitfil [10]

Answer:

Yes the water will be safe at the point of cooling water discharge

Explanation:

Power losses in plant= 350- 350×0.35=227.5MW

Rate of heat rejection to stream= 0.75× 227.5= 170.625MW

Rate of heat rejection= rate of flow of water× c × ΔT

170625000= 150000000× 4.186 × (Final temperature- 20)

Final temperature= 20.3 ◦C

The final temperature of stream will be 20.3 ◦C. Thechange is very small so the minnows will be able to handle this temperature.

7 0

3 years ago

A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second

Andreyy89

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.

4 0

3 years ago

java Write a program that simulates tossing a coin. Prompt the user for how many times to toss the coin. Code a method with no p

max2010maxim [7]

Answer:

The solution code is written in Java.

public class Main {
public static void main(String[] args) {
Scanner inNum = new Scanner(System.in);
System.out.print("Enter number of toss: ");
int num = inNum.nextInt();
for(int i=0; i < num; i++){
System.out.println(toss());
}
}
public static String toss(){
String option[] = {"heads", "tails"};
Random rand = new Random();
return option[rand.nextInt(2)];
}
}

Explanation:

Firstly, we create a function toss() with no parameter but will return a string (Line 14). Within the function body, create an option array with two elements, "heads" and "tails" (Line 15). Next create a Random object (Line 16) and use nextInt() method to get random value either 0 or 1. Please note we need to pass the value of 2 into nextInx() method to ensure the random value generated is either 0 or 1. We use this generate random value as an index of option array and return either "heads" or "tails" as output (Line 17).

In the main program, we create Scanner object and use it to prompt user to input an number for how many times to toss the coin (Line 6 - 7). Next, we use the input num to control how many times a for loop should run (Line 9). In each round of the loop, call the function toss() and print the output to terminal (Line 10).

4 0

3 years ago

Read 2 more answers

What is the composition of the two phases that form when a stream of 40% A, 39% B, and 21% C separates into two phases? Label th

irga5000 [103]

Answer:

vapor fraction = 0.4 and 0.08

Explanation:

At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

8 0

3 years ago

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

3 years ago