10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.
Kph is very common for vehicles:
10 km/10 min (60 min/hr) = 60 kph
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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<h3>
Answer:</h3>
225 meters
<h3>
Explanation:</h3>
Acceleration is the rate of change in velocity of an object in motion.
In our case we are given;
Acceleration, a = 2.0 m/s²
Time, t = 15 s
We are required to find the length of the slope;
Assuming the student started at rest, then the initial velocity, V₀ is Zero.
<h3>Step 1: Calculate the final velocity, Vf</h3>
Using the equation of linear motion;
Vf = V₀ + at
Therefore;
Vf = 0 + (2 × 15)
= 30 m/s
Thus, the final velocity of the student is 30 m/s
<h3>Step 2: Calculate the length (displacement) of the slope </h3>
Using the other equation of linear motion;
S = 0.5 at + V₀t
We can calculate the length, S of the slope
That is;
S = (0.5 × 2 × 15² ) - (0 × 15)
= 225 m
Therefore, the length of the slope is 225 m
Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s
then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9
Explanation:
We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.
We have Vr = (5.37m/s, 0m/s)
Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).
The velocity of the boat Vb will be:
Vb = (-c*sin(32). c*cos(32))
Then we should have that:
5.37 m/s - c*sin(32) = 0
c = (5.37/sin(32))m/s = 10.13 m/s
the velocity in the y-axis is:
10.13m/s*cos(32) = 8.59 m/s
So Vb = (-5.37m/s, 8.59 m/s)
the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.
V=wave velocity , <span>f= frequency, </span><span>λ=wavelength </span>
<span>Use it to find corresponding wavelengths for</span><span> f=28 Hz </span>
<span>λ= v/f= 337/28=12.036 m
</span>
<span>for f=4200 Hz </span>
<span>λ= v/f=337/4200= 0.08 m </span>
<span>So max. wavelength is 12.036 m and </span>
<span>Min Wavelength is 0.08 m </span>
<span>So the range is between .08 m and 12.036 m
</span>Hope this helps.