Answer:
3.14 × 10⁻⁴ m³ /s
Explanation:
The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.
Q = Area x velocity
Given:
Diameters of 3 sections of the pipe are given as
d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.
Speed in the first segment of the pipe is
v1 = 4 m/s.
From the equation of continuity the flow rate through different cross-sections remains the same.
Flow rate = Q = A1 v1 = A2 v2 = A3 v3.
Q = A1v1
=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s
Answer:
M g H / 2 = M g L / 2 initial potential energy of rod
I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod
M g L / 2 = 1/3 M L^2 * ω^2 / 2
g = 3 L ω^2
ω = (g / (3 L))^1/2
Answer:
Explanation:
According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.
Being 
and 
the masses of pucks a and b respectively, the initial momentum of the system is
Since b is initially at rest
After the collision and being 
and 
the respective velocities, the total momentum is
Both momentums are equal, thus
Solving for 
The initial kinetic energy can be found as (provided puck b is at rest)
The final kinetic energy is
The change of kinetic energy is
Answer:
C1 + C2 = 30 parallel connection
C1 * C2 / (C1 + C2) = 7.2 series connection
C1 * C2 = 7.2 * (C1 + C2) = 216
C2 + 216 / C2 = 30 using first equation
C2^2 + 216 = 30 C2
C2^2 - 30 C2 + 216 = 0
C2 = 12 or 18 solving the quadratic
Then C1 = 18 or 12
The first one is revolves
