Answer:
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
Explanation:
a) The mass flow rate through the nozzle can be calculated with the following equation:
Where:
: is the initial velocity = 20 m/s
: is the inlet area of the nozzle = 60 cm²
: is the density of entrance = 2.21 kg/m³
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
Therefore, the exit area of the nozzle is 23.6 cm².
I hope it helps you!
Answer:41.991ml
Explanation:
Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION
2 H+ + 2e → H2 REDUCTION
Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.
from the oxidation reaction:
1moles of oxygen requires 4moles of electrons to be discharged at the product
F=96500C/mol
Quantity of charge Q=It
=60*60*0.201A
Q=723.6C
Mole=Q/(F*mole ratio of electron)
Mole= 723.6/(4*96500)
Mole=((1809)/(965000))
M=0.0018746114
M1/M2=V1/V2
1/0.00187=22.4dm^3/V2
V2=22.4*0.00187
V2=0.04199129534dm^3
41.99129534ml
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = 
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
