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What is the period of a wave whose frequency is 2MHz?
The period is the reciprocal of the frequency. 1 / 2 million Hz = 500 ns or 0.5 us (microseconds).
What is the relationship between wavelength and frequency?
Wavelength, λ, and Frequency, f, are inversely proportional. Their product is a constant, the wave velocity. For lightwaves, their product is the speed of light, c: c = λ * f = 299,792,458 m/s, ~= 3.00 * 10**8 m/s, 0.300 m/ns, 30.0 cm/ns, ~= 186,000 mi/s, 11.811 in/ns, 0.984252 ft/ns; Where ns = nanoseconds, or 10**-9 seconds. Frequency, f, and Wavelength, λ, describe simple values that can be measured on a moving wave, if it contains a constant signal or, at least, a clearly repetitive waveform. So, as the wave moves past a reference point like a microphone, or radio receiver, find some repetition in the wave, measure the shortest interval of time, t, that marks the entire repetition. The frequency, f, is simply[1] the reciprocal of that time interval: f = 1 / t The wavelength is the measured distance between two identical areas of the waveform (like peaks, troughs, or blips), in adjacent copies of a repeating waveform. Whereas the interval measurements required one wave sensor, plus a fast timer, the wavelength measurements require two wave sensors, accurate distance measurements, but no timer. As the sensors are separated, the two signals will diverge until, at one wavelength separation, the two signals become identical again (difference is minimal). Sometimes, it is hard to be highly accurate about where, exactly, a single interval starts and stops. However, if you can flawlessly count a large number of repetitions, just do your best to start and stop the timer in the same place on the first and last repetition. Now you get to multiply your accuracy! Divide the total time, Δt, by the number of intervals, n, this average is an improved measurement of interval! t = Δt / n If you divide the count by the time interval, you get an improved frequency measurement. [1] Frequency is a counted number, n, of full waveform repetitions divided by the total elapsed time, Δt. f = n / Δt
How many ns are in 1 second?
The prefix nano (n) means 10-9 of something, in this case 1 ns are equivalent to 10-9 s. So 1 second is composed by 109 ns.
How many nanoseconds are in second?
1 ns (nanosecond) = 10-9s 1s / 10-9s = 109 109 = 1000 000 000 1 second = 1000 000 000 ns (nanosecond)
What you mean by 'pulse duration of 10 ns?
If that doesn't already make sense to you, I can't think of any explanation that would.
What is the period of a wave whose frequency is 2MHz?
The period is the reciprocal of the frequency. 1 / 2 million Hz = 500 ns or 0.5 us (microseconds).
What is the speed in terms of Mhz of a clock period of 1 ns?
A clock with a period of 1 ns has a frequency of 1 GHz, or 1000 MHz.
What is the period of a 1GHz sine wave?
The period of 1GHz is 1 ns. The waveform is irrelevant.
What is the period of a 15MHz sine wave?
The period of a 15MHz sine wave is 1 / 15MHz, or 0.066667 us, or 66 2/3 ns.
Give the clock out frequency and state time T of 8085 when the crystal frequencies is 5MHz?
The clock out frequency of an 8085 is one half the crystal frequency. The period of one T cycle is the inverse of the clock frequency. At a crystal frequency of 5MHz, the clock is 2.5MHz, and T is 400 ns.
What is the relationship between wavelength and frequency?
Wavelength, λ, and Frequency, f, are inversely proportional. Their product is a constant, the wave velocity. For lightwaves, their product is the speed of light, c: c = λ * f = 299,792,458 m/s, ~= 3.00 * 10**8 m/s, 0.300 m/ns, 30.0 cm/ns, ~= 186,000 mi/s, 11.811 in/ns, 0.984252 ft/ns; Where ns = nanoseconds, or 10**-9 seconds. Frequency, f, and Wavelength, λ, describe simple values that can be measured on a moving wave, if it contains a constant signal or, at least, a clearly repetitive waveform. So, as the wave moves past a reference point like a microphone, or radio receiver, find some repetition in the wave, measure the shortest interval of time, t, that marks the entire repetition. The frequency, f, is simply[1] the reciprocal of that time interval: f = 1 / t The wavelength is the measured distance between two identical areas of the waveform (like peaks, troughs, or blips), in adjacent copies of a repeating waveform. Whereas the interval measurements required one wave sensor, plus a fast timer, the wavelength measurements require two wave sensors, accurate distance measurements, but no timer. As the sensors are separated, the two signals will diverge until, at one wavelength separation, the two signals become identical again (difference is minimal). Sometimes, it is hard to be highly accurate about where, exactly, a single interval starts and stops. However, if you can flawlessly count a large number of repetitions, just do your best to start and stop the timer in the same place on the first and last repetition. Now you get to multiply your accuracy! Divide the total time, Δt, by the number of intervals, n, this average is an improved measurement of interval! t = Δt / n If you divide the count by the time interval, you get an improved frequency measurement. [1] Frequency is a counted number, n, of full waveform repetitions divided by the total elapsed time, Δt. f = n / Δt
Nanoseconds is the period of the cesium-133 radiation used to define the second?
1,087,827,757 ns
What is (-3) and divide(-2) plus 8?
Wavelength, λ, and Frequency, f, are inversely proportional. Their product is a constant, the wave velocity. For lightwaves, their product is the speed of light, c: c = λ * f = 299,792,458 m/s, ~= 3.00 * 10**8 m/s, 0.300 m/ns, 30.0 cm/ns, ~= 186,000 mi/s, 11.811 in/ns, 0.984252 ft/ns; Where ns = nanoseconds, or 10**-9 seconds. Frequency, f, and Wavelength, λ, describe simple values that can be measured on a moving wave, if it contains a constant signal or, at least, a clearly repetitive waveform. So, as the wave moves past a reference point like a microphone, or radio receiver, find some repetition in the wave, measure the shortest interval of time, t, that marks the entire repetition. The frequency, f, is simply[1] the reciprocal of that time interval: f = 1 / t The wavelength is the measured distance between two identical areas of the waveform (like peaks, troughs, or blips), in adjacent copies of a repeating waveform. Whereas the interval measurements required one wave sensor, plus a fast timer, the wavelength measurements require two wave sensors, accurate distance measurements, but no timer. As the sensors are separated, the two signals will diverge until, at one wavelength separation, the two signals become identical again (difference is minimal). Sometimes, it is hard to be highly accurate about where, exactly, a single interval starts and stops. However, if you can flawlessly count a large number of repetitions, just do your best to start and stop the timer in the same place on the first and last repetition. Now you get to multiply your accuracy! Divide the total time, Δt, by the number of intervals, n, this average is an improved measurement of interval! t = Δt / n If you divide the count by the time interval, you get an improved frequency measurement. [1] Frequency is a counted number, n, of full waveform repetitions divided by the total elapsed time, Δt. f = n / Δt
If clock frequency is 5 MHz how much time is required to execute an instruction of 18 T states?
At a clock frequency of 5 MHz (10 MHz crystal) the 8085 has a clock period of 200 ns. An instruction using 18 cycles would use 3.6 us. (Microseconds)This is for the case with no wait states. Each wait state adds 200 ns. Since an 18 cycle instruction has 5 memory accesses, one wait state per access would add 1 us to the execution time.
If the clock frequency is 5 MHz how much time is required to execute an instruction of 18 T states?
At a clock frequency of 5 MHz (10 MHz crystal) the 8085 has a clock period of 200 ns. An instruction using 18 cycles would use 3.6 us. (Microseconds)This is for the case with no wait states. Each wait state adds 200 ns. Since an 18 cycle instruction has 5 memory accesses, one wait state per access would add 1 us to the execution time.
What are the stator of poles?
In an Induction motor synchronous speed is inversely proportion to No. of poles if we increase No. of poles speed will decrease they are derived through formula as under P= 120*f/Ns Where P= No. of Poles f= Rated frequency Ns= Synchronous speed of flux.
What is the distance from Halifax NS and Sackville NS?
There are about 9.778 miles between Halifax, NS and Sackville NS.
