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Lubov Fominskaja [6]
3 years ago
15

The files provided in the code editor to the right contain syntax and/or logic errors. In each case, determine and fix the probl

em, remove all syntax and coding errors, and run the program to ensure it works properly.
Engineering
1 answer:
Yanka [14]3 years ago
3 0

Question Continuation

public class DebugOne3{

public static void main(String args){

System.out.print1n("Over the river");

System.out.pr1ntln("and through the woods");

system.out.println("to Grandmother's house we go");

}

}

Answer:

Line 2: Invalid Syntax for the main method. The main method takes the form

public static void main (String [] args) { }

or

public static void main (String args []) { }

Line 3: The syntax to print is wrong.

To print on a new line, make use of System.out.println(".."); not System.out.print1n();

Line 4:

To print on a new line, make use of System.out.println(".."); not System.out.pr1ntln();

Line 5:

The case of "system" is wrong.

The correct case is sentence case, "System.out.println" without the quotes

The correct program goes, this:

public class DebugOne3{

public static void main(String [] args){

System.out.println("Over the river");

System.out.println("and through the woods");

System.out.println("to Grandmother's house we go");

}

}

Explanation:

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Define cooling tower "range" as it applies to cooling towers.
Arlecino [84]

Answer:

The answer to the range of a cooling tower is the difference between the temperature of the water going in and the temperature of the water going out of the tower.

Explanation:

Cooling towers are used to cool a stream of water using the air of the environment where it's placed or installed. In theory, the lowest temperature the water can be cooled to would be the wet bulb temperature (temperature of the air going in the cooling tower). The range would be the difference (R=Ti-To) between the temperature of the hot water going in and the cooler water going out.

8 0
3 years ago
An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100
Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}

Where:

\sigma_{m} - Mean stress, measured in megapascals.

\sigma_{min} - Minimum stress, measured in megapascals.

\sigma_{max} - Maximum stress, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the mean stress is:

\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}

\sigma_{m} = 100\,MPa

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

\sigma_{a} = |\sigma_{max}-\sigma_{min}|

Where \sigma_{a} is the stress amplitude, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the stress amplitude is:

\sigma_{a} = |300\,MPa-(-100\,MPa)|

\sigma_{a} = 400\,MPa

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio (R) is the ratio of the stress amplitude to mean stress. That is:

R = \frac{\sigma_{a}}{\sigma_{m}}

If we know that \sigma_{m} = 100\,MPa and \sigma_{a} = 400\,MPa, the stress ratio is:

R = \frac{400\,MPa}{100\,MPa}

R = 4

The stress ratio of the aluminium part is 4.

3 0
3 years ago
Which of the following is the next “up-and-coming” digital publishing industry?
storchak [24]

Answer:

3D printing

Explanation:

The developments in technology, infrastructure, systems and technology mean that 3D printing will soon become a production technology. The market penetration of 3D printing would inevitably rise over time, with certain categories almost completely transitioning to 3D printing, such as digital publishing.

6 0
3 years ago
In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference be
MArishka [77]

Answer:

Hence the given statement is false.

Explanation:

For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.

The speed within the test section is decided by the planning of the tunnel.  

Thus by adjusting the pressure difference won't change the worth of test section velocity.

7 0
3 years ago
Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th
Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

PV=mRT

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

T_1 = 38\°C

T_2 = 14\°C

\eta = 97\%

\dot{v} = 510m^3/kg

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

PV = mRT

\dot{m} = \frac{PV}{RT}

\dot{m} = \frac{0.6626*10^{5}*510}{287*311}

\dot{m} = 37.85kg/min

Therfore the mass flow rate at which water condenses, then

\eta = \frac{\dot{m_v}}{\dot{m}}

Re-arrange to find \dot{m_v}

\dot{m_v} = \eta*\dot{m}

\dot{m_v} = 0.97*37.85

\dot{m_v} = 36.72 kg/min

PART B) Enthalpy is given by definition as,

H= H_a +H_v

Where,

H_a= Enthalpy of dry air

H_v= Enthalpy of water vapor

Replacing with our values we have that

H=m*0.0291(38-25)+2500m_v

H = 37.85*0.0291(38-25)-2500*36.72

H = 91814.318kJ/min

In the conversion system 1 ton is equal to 210kJ / min

H = 91814.318kJ/min(\frac{1ton}{210kJ/min})

H = 437.2tons

The cooling requeriment in tons of cooling is 437.2.

3 0
3 years ago
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