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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For a constant-velocity object, the average and instantaneous are the same. So the answer is no. It's like taking a running average of a string of numbers that are all the same number. The average is always the sum of the numbers divided by how many have accumulated, which will always equate to the repeated number.
The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>
Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:
is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following
Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:
638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>
Hi there!
We can begin by solving for the linear acceleration as we are given sufficient values to do so.
We can use the following equation:
vf = vi + at
Plug in given values:
4 = 9.7 + 4.4a
Solve for a:
a = -1.295 m/s²
We can use the following equation to convert from linear to angular acceleration:
a = αr
a/r = α
Thus:
-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.
Now, we can find the angular displacement using the following:
θ = ωit + 1/2αt²
We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:
v = ωr
v/r = ω
9.7/0.61 = 15.9 rad/sec
Plug into the equation:
θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad
