Elements in the same ____ have the same numbers of valence electrons?
2 answers:
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Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v =
in this expression the densities are constant
v = A √h
A =
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A =
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h =
Δh = Δv
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
= 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
= 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
Answer:
367.43 mm²
Explanation:
Given:
Flow rate, Q = 0.7 L/s
1000 L = 1 m³ = 10⁹ mm³
thus,
1 L = 10⁶ mm³
Therefore,
Q = 0.7 × 10⁶ mm³/s
Cross-sectional area at the top of the sprue = 750 mm²
Length of the sprue = 185 mm
Now,
Velocity =
where, g is the acceleration due to gravity = 9.81 m/s²
h is the height through which flow is taking place = 185 mm = 0.185
thus,
Velocity =
or
velocity = 1.9051 m/s = 1905.1 mm/s
Also,
Q = Area × Velocity
thus,
0.7 × 10⁶ = Area × 1905.1
or
Area = 367.43 mm²