Answer:
See below
Explanation:
<u>I will use 3 x 10^8 m/s for speed or wave</u>
speed = wavelength * frequency
3 x 10^8 = w * 7.34 x 10^2 <====== are you sure this isn't KILO Hz ?
w = <u>408719. 3 meters </u>
Answer:
Explanation:
The path length difference = extra distance traveled
The destructive interference condition is:

where m =0,1, 2,3........
So, ←
![\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}](https://tex.z-dn.net/?f=%5CDelta%20d%20%3D%20%28m%2B1%2F2%29%5Clamb%20da9%2Ftex%5D%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3Eso%20%3C%2Fstrong%3E%5Btex%5D%5CDelta%20d%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%7D)
⇒ λ = 2Δd = 2×10 = 20
Magnitude of displacement = 
Adding the squares gives displacement = 
Displacement =
≈ 724.7m
Line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration is force(10, 9.81).
<h3 /><h3>Line of code for force and acceleration</h3>
- In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies.
- Acceleration is a vector quantity (in that they have magnitude and direction).
- The direction of an object's acceleration is determined by the direction of the net force acting on it.
- Newton's Second Law states that the combined effect of two factors determines how much an item accelerates.
- The size of the net balance of all external forces acting on the object is, in accordance with the materials used to create it.
- It inversely proportional to its mass, whereas the magnitude of the net resultant force is directly proportional to the net force.
def force(mass, acceleration):
force_val = mass*acceleration
return force_val
10 is assigned to mass and 9.81 is assigned to acceleration
def force(10, 9.81)
So, Line of code will call force with a value of 10 for mass and a value of 9.81 for acceleration is force(10, 9.81).
Learn more about acceleration here:
brainly.com/question/460763
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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