Answer:
V = 280.15 V
Explanation:
" The complete question is attached with figure"
Given:
- The capacitance of the capacitor C = 10 nF
- The amount of mass attached to motor m = 4 grams
- The amount of distance it is to be lifted h = 1 cm
- Ignore all other losses in the system
Find:
- The voltage required to lift the mass m through distance h?
Solution:
- The conservation of energy for the entire system is written as:
Work_gravity = U_c
Where,
Work_gravity: Work done by gravity on mass m
U_c: The amount of energy stored in a capacitor
m*g*h = 0.5*C*V^2
V^2 = 2*m*g*h / C
V = sqrt ( 2*m*g*h / C )
Plug in the values:
V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )
V = sqrt ( 78,480)
V = 28.15 V
Answer:
5.118 m^3/hr
Explanation:
Given data:
viscosity of cell broth = 5cP
cake resistance = 1*1011 cm/g
dry basis per volume of filtrate = 20 g/liter
Diameter = 8m , Length = 12m
vacuum pressure = 80 kpa
cake formation time = 20 s
cycle time = 60 s
<u>Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter</u>
attached below is a detailed solution of the question
Hence The filtration rate in volumes/hr expected for the rotary vacuum filter
V' = ( ) * 1706.0670
= 5118.201 liters ≈ 5.118 m^3/hr
This question is incomplete, the complete question is;
the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate ( in m³ ) if the rectangular channel is 20 m wide.
Answer:
the flow rate is 32.549 m³/sec
Explanation:
Given that
y₁ = 0.3 m
y₂ = 1.2 m
β = 20 m
Now for Rectangular Channel, we know that;
2q²/g = y₁y₂( y₁ + y₂)
where g = 9.81 m/s²
and q = Q/β
so
2(Q/β)²/g = y₁y₂( y₁ + y₂)
we substitute our given values
2(Q/20)²/9.81 = 0.3 × 1.2( 0.3 + 1.2)
2(Q²/400)/9.81 = 0.36(1.5)
2(Q²/400) = 0.54 × 9.81
Q²/400 = 5.2974 / 2
Q²/400 = 2.6587
Q² = 1059.48
Q = √1059.48
Q = 32.549 m³/sec
Therefore the flow rate is 32.549 m³/sec