We have that the sea level pressure for Leh area is 1150mb mathematically given as
Ps= 1150 mb
<h3>
Sea level pressure</h3>
Question Parameters:
Ladakh is 800 mb.
<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m
increase in height with respect to sea level corresponds to 10 mb pressure,
Generally, for 3500m the pressure change will be 350 mb.
Therefore, here for the sea level <em>pressure</em> we need to add,
Ps=800+350
Ps= 1150 mb
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Answer:
Re = 1 10⁴
Explanation:
Reynolds number is
Re = ρ v D /μ
The units of each term are
ρ = [kg / m³]
v = [m / s]
D = [m]
μ = [Pa s]
The pressure
Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]
μ = [Pa s] = [kg / m s²] [s] = [kg / m s]
We substitute the units in the equation
Re = [kg / m³] [m / s] [m] / [kg / m s]
Re = [kg / m s] / [m s / kg]
RE = [ ]
Reynolds number is a scalar
Let's evaluate for the given point
Where the data for methane are:
viscosity μ = 11.2 10⁻⁶ Pa s
the density ρ = 0.656 kg / m³
D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m
Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶
Re = 1.19 10⁴
Answer:
Explanation:
She's correct but doesn't mean the wagon cannot put into motion. The force that she applied on the wagon, according to Newton's 2nd law, would have generated an acceleration, which translates into motion. The reaction force the wagon applies on her due to Newton's 3rd law, would not hinder its own motion.
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
