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
First, we need to establish the motion equations for the cart.
This leads us to:
vf = a * t + vo
By substituting the values, we have:
8.2 = a * (1.5) + (3.5)
Solving for acceleration gives:
a = (8.2-3.5) / (1.5)
a = 3.1 m / s ^ 2
Response:
The acceleration of the cart is:
a = 3.1 m / s ^ 2
Response:
The man's speed is 0.144 m/s
Explanation:
This exemplifies conservation of momentum.
The momentum of the ball prior to being caught must equal the momentum of the man-ball system after catching the ball.
Mass of the ball = 0.65 kg
Mass of the man = 54 kg
Speed of the ball = 12.1 m/s
The momentum of the ball before impact can be calculated as mass multiplied by velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After catching the ball, the momentum of the combined system is
(0.65 + 54)Vf = 54.65Vf
Where Vf denotes their final shared velocity.
Setting the initial momentum equal to the final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
U = 0, the initial vertical velocity
Ignoring air resistance, with g set to 9.8 m/s².
The duration, t, required for the pen to reach a vertical speed of 19.62 m/s can be calculated with
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s
Result: 2.0 s
