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.
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The elevation in reservoir at the rate of flow using is 03m/s is 114m.
The Reynolds range is the ratio of inertial forces to viscous forces. The Reynolds variety is a dimensionless variety used to categorize the fluids structures in which the impact of viscosity is crucial in controlling the velocities or the flow sample of a fluid.
The reason of the Reynolds number is to get a few experience of the relationship in fluid glide between inertial forces (this is those that maintain going by using Newton's first law – an item in motion stays in movement) and viscous forces, this is people who cause the fluid to come back to a forestall because of the viscosity of the fluid.
calculation,
Let L = 100 m pipe
L1 = 150 m pipe
H f = friction losses
Using Reynolds number, relative roughness, friction co- effiicients and friction losses
Substitute the value in equation
Z = 110= 0.48= 3.54
Z = 114m
Therefore water surface elevation at reservoir is 114 meter.
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Answer:
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.
Explanation:
Mass of object 1 , m₁ = 300 g = 0.3 kg
Mass of object 2 , m₂ = 400 g = 0.4 kg
Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s
Initial velocity of object 2 , v₂ = 3.00j m/s
Mass of composite = 0.7 kg
We need to find final velocity of composite.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s
Final momentum = 0.7 x v = 0.7v kgm/s
Comparing
1.5 i + 0.24 j = 0.7v
v = 2.14 i + 0.34 j
Magnitude of velocity
Direction,
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.
Answer:
15.7 m
Explanation:
The range (horizontal distance) of the projectile is determined only by its horizontal motion.
The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:
where
v = 12.0 m/s is the initial velocity
is the angle between the direction of v and the horizontal
Substituting,
We know that the projectile hits the ground in a time of
t = 2.08 s
so the horizontal distance covered is
