Answer:
T = 0.003 s
(Period is written as T)
Explanation:
Period = time it takes for one wave to pass (measured in seconds)
frequency = number of cycles that occur in 1 second
(measured in Hz / hertz / 1 second)
Period : T
frequency : f
So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period
T = 
f = 
to find the period (T) of this wave, we need to plug in the frequency (f) of 300
T = 
T = 0.00333333333
So, the period of a wave that has a frequency of 300 Hz is 0.003 s
[the period/T of this wave is 0.003 s]
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)
Answer:
Explanation:
When the pendulum falls freely the net acceleration due to gravity is zero.
As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.
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
