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Despeckle

This is a median filter. It replaces each pixel with the median value in its 3 x 3 neighborhood. This is a time consuming operation because, for each pixel in the selection, the nine pixels in the 3×3 neighborhood must be sorted and the center pixel replaced with the median value (the fifth). Median filters a good at removing salt and pepper noise. from imageJ wiki

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What is the work output by a 92 percent efficient wheel and axle when 75 J of work is input?

0.92 x 75 J = 69 joules How did you get that answer


The efficiency of a ramp is 75. If the amount of work input is 240 J what is the amount of useful work output?

The amount of useful work output can be calculated by multiplying the work input by the efficiency. In this case, 240 J * 0.75 = 180 J. Therefore, the useful work output is 180 J.


What is the efficiency of a pulley that requires a worker to do 250 j of work to lift an object that requires 170 j or work?

The efficiency of the pulley system is calculated by dividing the output work by the input work and multiplying by 100, so efficiency = (output work/input work) * 100. In this case, the output work is 170 J and the input work is 250 J, so the efficiency is (170/250) * 100 = 68%.


How much work did a system do in kJ when a system absorbs 650 J during a change and the value was -2.37?

The work done by the system can be calculated using the formula: Work = Force × Distance × cos(θ). Since work is given as -2.37 kJ, we convert this value to joules (1 kJ = 1000 J). So, the work done is -2.37 kJ × 1000 J/kJ = -2370 J. Therefore, the work done by the system when it absorbs 650 J of energy during a change is -2370 J + 650 J = -1720 J or -1.72 kJ.


What is the efficiency percentage of a mechanical device which requires 430 J of work to do 230 J of work in lifting a crate?

The efficiency of the mechanical device is calculated by dividing the output work (230 J) by the input work (430 J) and multiplying by 100. Therefore, the efficiency percentage would be (230 J / 430 J) * 100 = 53.5%.

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To effectively reduce noise in GIMP and enhance image quality, you can use the "Despeckle" filter or the "Selective Gaussian Blur" filter. Additionally, adjusting the brightness and contrast levels can help reduce noise. Experiment with these tools to find the best settings for your specific image.


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What is the work output by a 92 percent efficient wheel and axle when 75 J of work is input?

0.92 x 75 J = 69 joules How did you get that answer


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Java Swings has a lot of nice features and one of them is the ability to display images on its components. You can select any image that is available in the local system or somewhere on the internet and display it on Graphics objects. As of Java SE 1.4, reading an image is very simple. If the image is stored in a local file: String filename = "..."; Image image = ImageIO.read(new File(filename)); Otherwise, you can supply a URL: String urlname = "..."; Image image = ImageIO.read(new URL(urlname)); The read method throws an IOException if the image is not available. Now the variable image contains a reference to an object that encapsulates the image data. You can display the image with the drawImage method of the Graphics class. public void paintComponent(Graphics g) { . . . g.drawImage(image, x, y, null); } Sometimes, you may want to tile an image across a components canvas if it is much smaller than the component size. We can do the tiling in the paintComponent method. We first draw one copy of the image in the top-left corner and then use the copyArea call to copy it into the entire window using the code below: for (int i = 0; i * imageWidth <= getWidth(); i++) for (int j = 0; j * imageHeight <= getHeight(); j++) if (i + j > 0) g.copyArea(0, 0, imageWidth, imageHeight, i * imageWidth, j * imageHeight);