Answer:
1.8 m/s
Explanation:
Here is the complete question
The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.
What is the rate of discharge at point B?
Solution
Using Bernoulli's equation,
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³
P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²
P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)
(h₂ - h₁) = 5 m
Substituting the values of the variables, we have
700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)
36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)
- 13000 Pa = 500 kg/m³(v₂² - v₁²)
(v₂² - v₁²) = - 13000 Pa/500 kg/m³
(v₂² - v₁²) = -26 m²/s²
By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B
πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m
d₁²v₁ = d₂²v₂
v₁ = v₂(d₂/d₁)²
= v₂(0.15/0.05)²
= v₂(3)²
= 9v₂
So, substituting v₁ = 9v₂ into the above equation, we have
(v₂² - v₁²) = -26 m²/s²
v₂² - 9v₂² = -26 m²/s²
- 8v₂² = -26 m²/s²
v₂² = -26 m²/s² ÷ (-8)
v₂² = 3.25 m²/s²
taking square root of both sides,
v₂ = √(3.25 m²/s²)
= 1.8 m/s
So, the rate of discharge at point B is 1.8 m/s
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25
above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip the
change in the thermal energy of the projectile and air is:
Answer: 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ