Answer:
Explanation:
The concept of Hooke's law was applied as it relates to deformation.
The detailed steps and appropriate substitution is as shown in the attached file.
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Answer:
h=32.1 km
Explanation:
<em>solution:</em>
using newton law of gravitational attraction and newton second law:
r= distance between two masses
at sea level
a=g
.............................(1)
.........................(2)
by substituting (2) and (1) acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)
so the weight of the object at a distance r from the centre of the earth (W=ma)
W=mg(Re^2/r^2)..........(3)
h the height above the surface of the earth: r=Re+h
putting the value of r in eq (3)
W=mg(Re/Re+h)^2
W=0.99 mg
solving for height h:
h=Re(1/√0.99)-(1))
h=32.1 km
Answer:
For the Top Side
- Strain ε = 0.00021739
- Elongation is 0.00260868 cm
For The Right side
- Strain ε = 0.00021739
-Elongation is 0.00347826 cm
Explanation:
Given the data in the question;
Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m
Thickness = 5 mm = 0.005 m
Force to the Top F = 15 kN = 15000 Newton
Force to the right F = 20 kN = 20000 Newton
elastic modulus, E = 115 GPa = 115 × 10⁹ pascal
Now, For the Top Side;
- Strain = σ/E = F / AE
we substitute
= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )
= 15000 / 69000000
Strain ε = 0.00021739
- Elongation
Δl = ε × l
we substitute
Δl = 0.00021739 × 12 cm
Δl = 0.00260868 cm
Hence, Elongation is 0.00260868 cm
For The Right side
- Strain = σ/E = F / AE
we substitute
Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )
= 20000 / 69000000
Strain ε = 0.000289855
- Elongation
Δl = ε × l
we substitute
Δl = 0.000289855× 12 cm
Δl = 0.00347826 cm
Hence, Elongation is 0.00347826 cm
Answer:
The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.
Explanation:
Data
<u>kLa</u> = 0.17/s
<u>Solubility of oxygen</u> = 8 × 10^-3 kg / m^3
<u>The maximum specific oxygen uptake rate </u>= 4 mmol O2 / g h.
<u>Concentration of oxygen</u> = 0.5 × 10^-3 kg/ m^3
<u>**The maximum cell density</u> = 50 g/l
___________________
The calculated maximum cell concentration:
xmax= kLa · CAL*/ qo
CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate
Replacing the data given
xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 4 mmol O2 / g h
4 mmol O2 / g h to kg O2/ g s
= 3.56 x 10^-3 kg O2/ g s
So then,
xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 3.56 x 10^-3 o kg O2/ g s
xmax= 3. 8 x 10^4 g/ m^3 = 38 g/l
_____________________