The correct answer for this question is this one:
<span>A snowstorm was predicted in Chicago. The possible upper air temperature, surface temperature, and air pressure of Chicago on that day. Normal atmospheric pressure is 29.9 inches of mercury. </span><em>I'm pretty sure the answer is 40 for upper air, 29 for surface temp, and 30 for air pressure. </em>Hope this helps answer your question and have a nice day ahead.
The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
<h3 /><h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
Pressure is found as the product of the density,acceleraton due to gravity and the height.
P₁=ρ₁gh₁
P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m
P₁=24014.88 N/m²
P₂=ρ₂gh₂
P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m
P₂=196.2 N/m²
P₃=ρ₃gh₃
P₃=850 kg/m³×9.81 (m/s²)×0.25
P₃=2084.625 N/m²
Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
To learn more about the pressure refer to the link;
brainly.com/question/356585
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1. Amperes, is the SI unit (also a fundamental unit) responsible for current.
2. 
Δq over Δt technically
Rearrange for Δq
I x Δt = Δq
1.5mA x 5 = Δq
Δq = 0.0075
Divide this by the fundamental charge "e"
Electrons: 0.0075 / 1.60 x 10^-19
Electrons: 4.6875 x 10^16 or 4.7 x 10^16
3. So we know that the end resistances will be equal so:
ρ = RA/L
ρL = RA
ρL/A = R
Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper
We are looking for L2 so we can isolate using algebra to get:
If you fill in those values you get 0.0205
or 2.05 cm
The answer is; refraction.
When waves, including light, pass through two media with different densities, the waves bend at the transition between the two media due to differences in speed and wavelength. Convex lenses bend (refract) light and concentrate it at one point called the focal point.
<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.
Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up?
Barbara is going due south at 5.9 m/s, so she's at (0,-5.9)
Neil is going due west at 1.4 m/s, so he's at (-1.4,0)
Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this.
So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0).
The magnitude of Neil's velocity as seen by Barbara is
sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s
The angle of his vector relative to due west will be
atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees
So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>
