The correct formula to calculate TCIR (Total Case Incident Rate) is: (Number of recordable cases) / (Total hours worked by all employees) * 200,000. This formula is used to measure the number of recordable cases per 100 full-time equivalent employees in a given period.
You are legally required to record and OSHA recordable case.
A heart attack may be an OSHA Recordable event, but it may not, depending on the circumstances. It it is thought to be the result of one's work assignment, it would be Recordable. But events like heart attacks can happen at work without being work-related, so some (many) are not OSHA recordable. A knowledgeable person would have to determine that on a case by case basis.
Since the speed of the wave is equal to the wavelength times the frequency, all you need to do is divide the speed by the wavelength in this case.
E = hf The energy per photon is equal to Planck's constant times the frequency, in this case 6.62606957×10−34 x 107.3x106
To calculate the lowest frequency that can be generated using mode 1, we need to divide the crystal frequency by two. So, the lowest frequency that can be generated in this case would be 11.0592 MHz divided by 2, which is equal to 5.5296 MHz.
In that case, it will have a low frequency.
ave speed does not care if you have stopovers or not. ave speed = total distance / total time
Not enough information. You can use the equation speed = frequency x wavelength, but in this case, you don't have enough data to calculate the speed.
Let's take an example. Suppose that the intermediate frequency is 10,7 MHz (FM). The local oscillator works on 110,7 MHz. First case: You receive a signal of 100 MHz, the mixer will generate a frequency of 110,7 + 100 = 210,7 MHz, which will be rejected by the band-pass filter. The difference of the two frequencies is 110,7 - 100 = 10,7 MHz (desired one). Second case: You receive a signal of 121,4 MHz. The sum of that frequency and the local oscillator is 232,1 MHz, which will be rejected. The difference is 121,4 - 110,7 = 10,7 MHz. So the image frequency in that case is going to be 121,4 MHz.
You can't do this unless you know more detail on the frequency modulator, like its modulation index, for example or the frequency deviation. For the amplitude modulator, it is easy. Bandwidth is twice the signal frequency so in this case the bandwidth is 178kHz for AM.
In this case, the frequency of a wave emitted by one person would increase (be perceived as having a higher frequency) by the other.In this case, the frequency of a wave emitted by one person would increase (be perceived as having a higher frequency) by the other.In this case, the frequency of a wave emitted by one person would increase (be perceived as having a higher frequency) by the other.In this case, the frequency of a wave emitted by one person would increase (be perceived as having a higher frequency) by the other.