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3 years ago

The volume of microbial culture is observed to increase according to the formula

Engineering

1 answer:

saw5 [17]3 years ago

3 0

The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

We are given that;

Volume of microbial culture is observed to increase according to the formula;

V = e^(t)

where;

t is in seconds

V is in m³

We want to now express V in in³ and t in minutes.

Now, from conversions;

1 m³ = 61024 in³

Also; 1 second = 1/60 minutes

according to formula for exponential decay, we know that;

V = ae^(bt)

Thus, we have;

61024V = ae^(¹/₆₀b(t(h))

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

Read more about subject of formula at; brainly.com/question/790938

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What organization which fire codes

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Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ F$

Change in temperature, $\Delta T = T_2-T_1$

$\Delta T = -2.43-87.745$

$\Delta T=-90.175^\circ F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

$=11.401+0.32198(82.898)$

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2 years ago

Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high

Vinvika [58]

Answer:

e.Fire resistance,Inexpensive,Non-toxic.

Explanation:

Desirable hydraulic property of fluid as follows

1. Good chemical and environment stability

2. Low density

3. Ideal viscosity

4. Fire resistance

5. Better heat dissipation

6. Low flammability

7. Good lubrication capability

8. Low volatility

9. Foam resistance

10. Non-toxic

11. Inexpensive

12. Demulsibility

13. Incompressibility

So our option e is right.

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3 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

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Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

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$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

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Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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Answer:

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