<span>To begin, the formula for finding frequency when wavelength is known is "f = c / w" when c is the constant velocity (3 * 10^8 m/s). To convert the wavelength into a common form (m/s), it will have to be multiplied by 10^-2. This leaves the equation as "f = 3.0 * 10^8 / (2.4 * 10^-5 * 10^-2), or 2.4 * 10^-7. This gives 1.25 * 10^15 m/s as the frequency.</span>
Answer:
The sediment deposited by glaciers is called Glacial deposition.
Answer:
D. Meters/Seconds
Explanation:
The time period of a wave is measured in seconds.
A typical wave involves both time and distance. Consider a sound wave, which is basically a periodic modulation of the local air pressure. We "hear" the sound because our ears respond to the variations of pressure.
The most common metric of a sound wave is frequency. This is the rate at which the change in pressure occurs, and is measured in cycles per second, formally known as "hertz". The period is the inverse of frequency andl has the units of seconds per cycle, commonly stated simply as seconds.
Electric energy??? there isn't much information from the question but I can infer it's electricity.
Answer:
The net emissions rate of sulfur is 1861 lb/hr
Explanation:
Given that:
The power or the power plant = 750 MWe
Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:
= 3988126.8 J
= 3.99 MJ
Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.
i.e.
The mass of the coal that is burned per sec 
The mass of the coal that is burned per sec = 187.97 lb/s
The mass of sulfur burned 
= 2.067 lb/s
To hour; we have:
= 7444 lb/hr
However, If a scrubber with 75% removal efficiency is utilized,
Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr
= 0.25 × 7444 lb/hr
= 1861 lb/hr
Hence, the net emissions rate of sulfur is 1861 lb/hr
