the waveforms below represent the inputs to a s-r flip-flop. ignoring the present state value, during which time interval(s) will the q output of the flip-flop be high?

Answers

Answer 1

The q output of the flip-flop will be high during the time interval between 2 and 3.

The S-R flip-flop has two inputs, S (set) and R (reset), and two outputs, Q and Q'. When S is high and R is low, the Q output is set to high, and when S is low and R is high, the Q output is reset to low. In this case, the waveform for the S input is high between 2 and 3, while the waveform for the R input is low throughout the duration.

Therefore, during the time interval between 2 and 3, the S input is high and the R input is low, so the Q output will be set to high. During all other time intervals, either the S input is low or the R input is high, so the Q output will remain low.

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Related Questions

You are troubleshooting an application problem and want to eliminate faulty memory as a source of the problem. Which command do you use?Mdsched.exe. Mds.chace. Mds.con

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If you are troubleshooting an application problem and suspect that faulty memory may be the cause of the issue, you can use the "mdsched.exe" command to check for any memory problems.

"mdsched.exe" command runs the Windows Memory Diagnostic tool, which will test your computer's memory for any errors or issues. Once the test is complete, it will provide you with a report that you can use to determine whether faulty memory was indeed the source of the problem. It is important to eliminate faulty memory as a possible cause before moving on to other troubleshooting steps, as memory issues can often be the root cause of many application problems.

To troubleshoot an application problem and eliminate faulty memory as a source of the issue, you should use the command "mdsched.exe". This is done as follows:
1. Open the Run dialog box by pressing the Windows key + R.
2. Type "mdsched.exe" into the dialog box and hit Enter.
3. The Windows Memory Diagnostic tool will open, offering options to restart now and check for problems or check for problems the next time you start your computer.
4. Choose the appropriate option to run the memory diagnostic test.

This command, mdsched.exe, will run the Windows Memory Diagnostic tool, which is designed to detect and diagnose any issues with your computer's memory. By using this tool, you can confirm whether or not faulty memory is contributing to your application problem.

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Hi! To eliminate faulty memory as a source of an application problem, you should use the command "mdsched.exe".

This command launches the Windows Memory Diagnostic tool, which checks your computer's memory for any issues that might be causing the problem with your application.

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a parallel rlc circuit contains a resistor r = 1 ω and an inductor l = 2 h. select the value of the capacitor so that the circuit is critically damped

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To design a critically damped parallel RLC circuit with a resistor R = 1Ω and an inductor L = 2H, you need to select the value of the capacitor C according to the following formula: C = 1 / (4 * R * L) Plug in the values for R and L: C = 1 / (4 * 1 * 2) C = 1 / 8 So, you need to select a capacitor with a value of 1/8 F (0.125 F) for the circuit to be critically damped.

To calculate the value of the capacitor required to make the parallel RLC circuit critically damped, we need to use the formula for the damping ratio, which is given by: ζ = R / (2√(L/C)) where R is the resistance, L is the inductance, C is the capacitance, and ζ is the damping ratio. For critically damped behavior, ζ = 1, which means: 1 = R / (2√(L/C)) Substituting the given values of R = 1 Ω and L = 2 H, we get: 1 = 1 / (2√(2/C)) Squaring both sides and rearranging, we get: C = 8/9 F Therefore, the value of the capacitor required to make the parallel RLC circuit critically damped is 8/9 F.

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To make a parallel RLC circuit critically damped, the value of the capacitor should be chosen so that the damping factor is equal to 1. In a parallel RLC circuit, the damping factor can be calculated using the formula:

damping factor = R / (2 * √(L * C))

Given that R = 1 Ω and L = 2 H, we can rearrange the formula to find the value of the capacitor (C):

C = (R^2) / (4 * L)

Plugging in the values, we get:

C = (1^2) / (4 * 2) = 1 / 8

Therefore, the value of the capacitor needed for the circuit to be critically damped is C = 1/8 F (farads).

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The Weld Center Vertices option is available for which of the
following Fill Hole Mode settings?

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As a general overview, the term "Weld Center Vertices" typically refers to a feature or option that is used in 3D modeling or computer-aided design (CAD) software.

What is the feature used for?

This feature is usually used in conjunction with a "Fill Hole" mode, which is a tool that is used to fill in holes or gaps in 3D models.

When the "Weld Center Vertices" option is enabled in a "Fill Hole" mode, the software will attempt to connect the vertices or points around the hole by creating a new surface or face that is centered on the vertices. This can be useful for creating a more uniform and seamless 3D model, particularly when dealing with complex shapes or irregular surfaces.

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