Nitrogen oxides play a critical role in photochemical smog. They give the smog its yellowish-brown hue. Indoor residential appliances like gas stoves and gas or wood heaters can be significant emitters of nitrogen oxides in poorly ventilated environments.
- Nitrogen dioxide (NO₂), ozone (O₃), peroxyacetyl nitrate (PAN), and chemical compounds with the -CHO group are the main harmful elements of photochemical smog (aldehydes). If present in high enough amounts, PAN and aldehydes can harm plants and irritate the eyes.
- The greatest sources of emissions are power plants, heavy construction equipment driven by diesel, other moveable engines, and industrial boilers. Cars, trucks, and buses are next in line.
Therefore , on conclusion i.e. two gases with molecules consisting of nitrogen and oxygen atoms are nitric oxide (NO) and nitrogen dioxide (NO₂). These nitrogen oxides play a part in the development of smog and acid rain, adding to the issue of air pollution.
To know more about photochemical smog
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Answer:
E = 31.329 N/C.
Explanation:
The differential electric field 
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where 
is the radius of curvature.
Now
,
where 
is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find 
and 
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
The velocity is given by:
V = √(Vx²+Vy²)
V = velocity, Vx = horizontal velocity, Vy = vertical velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for V:
V = √(6²+12²)
V = 13.42m/s
Now find the direction:
θ = tan⁻¹(Vy/Vx)
θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for θ:
θ = tan⁻¹(12/6)
θ = 63.4°
The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.
Answer:28m/s
Explanation:I got one point so yeah hope this help
Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turn of the coil, N = 200 turn
area of the coil, A = 300 cm² = 0.03 m²
magnitude of magnetic field, B = 30 mT = 0.03 T
maximum value of induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
E = NBA(2πf)
where;
f is the frequency of the coil
Therefore, the frequency of the coil is 7.07 Hz
