A ball is thrown straight upward and rises to a maximum height of 18.0 m
above its launch point. at what height above its launch point has the
speed of the ball decreased to one-half of its initial value?
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kinematic equation
v^2=u^2-2as; v0=,s=18,a=-10m/s/s
u=root 360
(u/2)^2=u^2-2as
=>s=13.5m
Answer:
0.0072 m³/s
Explanation:
Using Bernoulli's law
P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal
1/2ρv₂² - 1/2ρv₁² = P₁ - P₂
flow rate is constant
A₁v₁ = A₂v₂
A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²
A₂ = πr₂² = π (0.0225)² = 0.00159 m²
v₁ = (A₂ / A₁)v₂
v₁ = (0.00159 m²/ 0.0028278 m²) v₂ = 0.562 v₂
substitute v₁ into the Bernoulli's equation
1/2ρv₂² - 1/2ρv₁² = P₁ - P₂
500 ( 1 - 0.3161 ) v₂² = (31.0 - 24 ) × 10³ Pa
341.924 v₂² = 7000
v₂² = 20.472
v₂ = √ 20.472 = 4.525 m/s
volume follow rate = 0.00159 m² × 4.525 m/s = 0.0072 m³/s
There will be four unpaired electrons
The metal complex is [FeX₆]³⁻
X being the halogen ligand
X = F, CL, Br, and I
The oxidation of metal state is +3
The ground state configuration is
₂₆Fe =Is² 2s²2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Metal, Fe(III) ion electron configures
₂₆Fe³⁺ = Is2 2s² 2p⁶ 3s² 3p⁶ 3d⁵
The answer is 68 F. i hope this helps
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