The wavelength of the wave can be found using the formula: wavelength = wave speed / frequency. Plugging in the values: wavelength = 18 m/s / 5 Hz = 3.6 meters.
The speed of a wave can be calculated by multiplying its wavelength by its frequency. In this case, the speed of the wave would be 10 m/s (2m * 5Hz).
Frequency = Velocity / Wavelength = 100 m/s / 20 m = 5 s-1 or 5 Hz.
It will take 0.2 seconds to generate one complete wave vibration with a frequency of 5Hz because frequency is the number of cycles per second, so the time period of one cycle can be calculated as 1/5 = 0.2 seconds.
5
fiveFive hertz is five cycles per second.
The speed of a wave can be calculated by multiplying its wavelength by its frequency. In this case, the speed of the wave would be 10 m/s (2m * 5Hz).
I've got no idea what a "5 cycle wavelength" is. However, I would just apply this formula: v = fλ, where v is the velocity (speed in m/s) of the wave, f is the frequency (in hertz), and λ is the wavelength (in m).
The answer is in the question! 5 Hz Also, a wavelength cannot be 5 cycles - wrong units.
Frequency = Velocity / Wavelength = 100 m/s / 20 m = 5 s-1 or 5 Hz.
velocity is equal to frequency times wavelength. You have velocity and frequency given so wl = v/f. The wavelength is 3/5m, or 6cm.
It will take 0.2 seconds to generate one complete wave vibration with a frequency of 5Hz because frequency is the number of cycles per second, so the time period of one cycle can be calculated as 1/5 = 0.2 seconds.
5
fiveFive hertz is five cycles per second.
An object vibrating at 5Hz is completing 5 cycles of vibrations in one second. This frequency is relatively low and would result in slow oscillations. Examples include the low hum of a refrigerator or the gentle swaying of a pendulum.
The frequency of infrared radiation ranges from about 300 GHz to 400 THz. Infrared radiation falls between visible light and microwaves on the electromagnetic spectrum. The specific frequency of infrared light depends on its wavelength.
I would suggest what you mean is to increase speed droop sensitivity. That mean you reduce speed droop percentage set point. I will answer primary impact only assuming your plant is grid connected. Assuming your current speed droop percentage set point is 10% and you have a 100MW hydro turbine generator serving a 50-Hz grid system. Therefore your droop response due to system frequency deviation is 100MW for every 5Hz (10% of your system nominal frequency of 50 Hz), or +/- 20MW/Hz or 2MW/0.1Hz. What this number does? It will be used by your governor to compensate the load of your hydro turbine as system frequency deviates from the reference frequency. As system frequency increases by 0.1Hz, the governor will subtract generator output by 2MW from the current load set point. If system frequency decreases by 0.1Hz then the governor will add 2MW to the current load set point. This is done automatically. Now you want to increase its sensitivity of your governor from this set point to 5%. You are actually doubling the response. That means the new load response will be +4MW/0.1Hz if unit system frequency reduces from the reference frequency and -4MW/0.1Hz if system frequency increases above the reference frequency. If you increase the sensitivity further to 2.5, then the new response will be +/-8MW/0.1Hz.
if we take the signal as voice of any people the changing in frecuncy mean the voice will change .. on other hand when we speek about low frecuncy that men we speek about DC when we speek about hi frecuny we speek about AC . the difrent in the rang of frecuny change the circuit analysis method .. ..