Answer:
0.0406 m/s
Explanation:
Given:
Diameter of the tube, D = 25 mm = 0.025 m
cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²
Mass flow rate = 0.01 kg/s
Now,
the mass flow rate is given as:
mass flow rate = ρAV
where,
ρ is the density of the water = 1000 kg/m³
A is the area of cross-section of the pipe
V is the average velocity through the pipe
thus,
0.01 = 1000 × 4.9 × 10⁻⁴ × V
or
V = 0.0203 m/s
also,
Reynold's number, Re = 
where,
ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s
thus,
Re = 
or
Re = 611.39 < 2000
thus,
the flow is laminar
hence,
the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s
or
maximum velocity = 0.0406 m/s
Answer:
Explanation:
a) On the verge of tipping over, reaction acts at the corner A
When slippage occurs,
Block moves w/ const. velocity equilibrium
Three-force member: reaction at A must pass through B
tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )
b) When slippage occurs,
Block moves w/ const. velocity equilibrium
Three-force member: reaction at C must pass through G
k tanθ µ =
tan x/ H/2 , x H/2
Answer:
the two defects of a simple cell are:
1. Polarization
2. Local action
Explanation:
He would work on the thing like in the method you work on your question.
Answer:
The speed at point B is 5.33 m/s
The normal force at point B is 694 N
Explanation:
The length of the spring when the collar is in point A is equal to:
The length in point B is:
lB=0.2+0.2=0.4 m
The equation of conservation of energy is:
(eq. 1)
Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2
in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2
Replacing in eq. 1:
Replacing values and clearing vB:
vB = 5.33 m/s
The balance forces acting in point B is:
Fc-NB-Fs=0
Replacing values and clearing NB:
NB = 694 N
