Which is not a noun

Document the XSS stored exploit script: Use the View Source feature of the web page and create a screenshot of the few lines cod

Natali [406]

Answer:

Hold on let me ask my brother

Explanation:

5 0

3 years ago

CHALLENGE ACTIVITY 1.4.1: Basic syntax errors. Type the statements. Then, correct the one syntax error in each statement. Hints:

Lunna [17]

Answer:

Lets check each statement for the errors.

Explanation:

System.out.println("Predictions are hard.");

This statement has no syntax errors. When this statement is executed the following line will be displayed:

Predictions are hard.

System.out.print("Especially ');

This statement is missing the closing quotation mark. A single quotation mark is placed instead of double quotation mark in the statement.

The following error message will be displayed when this program statement will be compiled:

Main.java:15: error: unclosed string literal

String literals use double quotes. So to correct this syntax error, the statement should be changed as follows:

System.out.print("Especially");

The output of this corrected line is as following:

Especially

System.out.println("about the future.").

In this line a period . is placed at the end of the statement instead of a semicolon ; but according to the syntax rules statements should end in semicolons.

The error message displayed when this line is compiled is as following:

Main.java:15: error: ; expected

Main.java:15: error: not a statement

So in order to correct this syntax error the statement should be changed as following:

System.out.println("about the future.");

The output of this corrected line is as following:

about the future

System.out.println("Num is: " - userName);

There is a syntax error in this statement because of - symbol used instead of +

+ symbol is used to join together a variable and a value a variable and another variable in a single print statement.

The error message displayed when this line is compiled is as following:

Main.java:13: error: bad operand types for binary operator '-'

So in order to correct this syntax error the statement should be changed as following:

System.out.println("Num is: " + userName);

This line will print two things one is the string Num is and the other is the value stored in userName variable.

So let userName= 5 then the output of this corrected line is as following:

Num is: 5

8 0

3 years ago

A) Define the term magnetic flux

Radda [10]

Answer:

Magnetic flux is a measurement of the total magnetic field which passes through a given area. It is a useful tool for helping describe the effects of the magnetic force on something occupying a given area.

3 0

2 years ago

Read 2 more answers

A mass of 1.3 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com

aivan3 [116]

Answer:

The magnitude of work input is 178.79 kJ

Explanation:

The process is an isothermal compression process because the temperature inside the cylinder is constant.

W = nRT ln(P1/P2)

n is the number of moles of air in the cylinder = mass/MW = 1.3/29 = 0.045 kgmol

R is gas constant = 8.314 kJ/kgmol.K

T is temperature of air = 24 °C = 24+273 = 297 K

P1 is intial pressure of air = 120 kPa

P2 is final pressure of air = 600 kPa

W = 0.045×8.314×297 ln(120/600) = 111.11661 ln 0.2 = 111.11661 × -1.609 = -178.79 kJ

6 0

3 years ago

7.41. A 10.0-m3 tank contains steam at 275°C and 15.0 bar. The tank and its contents are cooled until the pressure drops to 1.8

Nat2105 [25]

Answer:

A) 127923.08KJ

B) 131.347°C

C) 51.083kg

Explanation:

A) This will be done in steps

STEP1: CALCULATE THE PROPERTIES OF THE STEAM IN THE INITIAL STATE:

The first stage is a superheat, because the steam is dry. Using superheat steam table. At

T₁ = 275 °C and p₁ = 15 bar

From steam table for superheated steam you get the following properties in that state:

specific volume v₁ = 0.150285m²/kg

specific enthalpy h₁ = 2978.67kJ/kg

Because the specific volume equals volume of the vessel divided by mass of water, we have;

v₁ = V/m

you can compute the mass of water vapor enclosed in the vessel:

m = V/v₁ = 10m³ / 0.150285m²/kg = 66.54 kg

STEP 2: CALCULATE THE PROPERTIES OF THE STEAM AT THE FINALE STATE:

In final state there is wet steam at

p₂ = 1.8 bar

Using the steam table for saturated steam, because the steam is wet, you get:

Saturated steam temperature T₂ = 131.347 °C

specific volume

liquid water vl₂ = 1.01071×10⁻³ m³/kg

water vapor vg₂ = 0.643419m³/kg

specific enthalpy

liquid water hl₂ = 552.141kJ/kg

water vapor hg₂ = 2721.93 kJ/kg

STEP 3: CALCULATE THE DRYNESS FRACTION:

Specific volume and enthalpy of the gas mixture are given by:

v₂ = vl₂ + (vg₂ - vl₂)∙x

h₂ = hl₂ + (hg₂ - hl₂)∙x

where x is the dryness (or vapor quality), which is the mass fraction of water vapor in the mixture:

x = m_vapor/m

During the cooling process both volume of the tank and mass of water enclosed does not change. Hence the specific volume of water is the same

v₁ = v₂

v₁ = vl₂ + (vg₂ - vl₂)∙x

So the dryness fraction in final state is:

x = (v₁ - vl₂) / (vg₂ - vl₂)

= (0.150285 m²/kg - 1.071×10⁻³ m³/kg) / (0.643419 m³/kg - 1.071×10⁻³ m³/kg)

= 0.232295

STEP 4: CALCULATE THE SPECIFIC ENTHALPY OF THE WET STEAM:

h₂ = hl₂ + (hg₂ - hl₂)∙x

= 552.141kJ/kg + (2721.93kJ/kg - 552.141kJ/kg) × 0.2322945

= 1056.171kJ/kg

STEP 5: CALCULATE THE HEAT REMOVED FROM THE TANK.

The heat removed from the tank equals mass water times change in specific enthalpy:

Q = m∙∆h = m∙(h₁ - h₂) =

= 66.54kg × (2978.67 kJ/kg - 1056.171kJ/kg)

= 127923.08kJ

Therefore the heat removed from the tank is 127923.08kJ

B) From the saturated steam table, the temperature of the wet steam is;

T₂ = 131.347°C

C) From the calculation above, the fraction of water vapor is x. So the fraction of liquid water is (1 - x).

Therefore the mass of liquid water which has condensed form the super hated steam in initial state will be;

ml = (1 - x)∙m

= (1 - 0.232295) × 66.54kg

= 51.083 kg

3 0

3 years ago