Answer:
a) v = 1.19 m/s, b) P₁ = 0.922 x 10⁵ Pa
Explanation:
1) We will apply the fluid continuity equation
Q = A v
The area of a circle is
A = π r² = π d²/4
v = Q / A = Q 4 / π d²
v = 0.006 4/π 0.08²
v = 1.19 m/s
2) Apply Bernoulli's equation, considering point 1 as the bladder and point 2 as where the urine exits
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
The problem states that
P₂ = 1.0013 x 10⁵ Pa
v₁ = 0
y₁ = 1 m
y₂ = 0
Density of water (ρ) = 1000 kg/m³
P₁ + ρ y₁ = P₂ + ½ ρ v₂²
P₁ = P₂ + ½ ρ v₂² - ρ g y₁
P₁ = 1.013 x 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 x 10⁵ + 708.5 - 9800
P₁ = 92208.5 Pa
P₁ = 0.922 x 10⁵ Pa
Answer:
v = 54.2 m/s
Explanation:
We can utilize conservation of energy to solve this issue.
Initial condition Higher
Em₀ = U = m g h
Final condition. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ (2gh)
Now let's perform the calculation
v = √ (2 * 9.8 * 150)
v = 54.2 m/s
E_total = 5.8 x 10⁴ N/C
Explanation: To determine the electric field at specified points, we must calculate the vectors individually for each charge and sum them. The electric field caused by each charged conductive sheet can be derived via Gauss's law with the understanding of scalar products between the electric field and relevant surfaces.
If the products have three nitrogen atoms, the reactants must have had the same quantity, as mass is conserved in a chemical reaction.
