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LenaWriter [7]
3 years ago
5

Which of the following leaf modifications is beneficial in a low-light environment, e.g., the understory of a rainforest? (More

than one choice may apply.)
(A) Iridescent and reflective.
(B) Undersides are red from high levels of anthocyanin pigments.
(C) Covered in hairs.
(D) Have a LARGE surface area.
(E) Have very SMALL surface area.
(F) Sharp (nonphotosynthetic) spines.
Engineering
1 answer:
Mnenie [13.5K]3 years ago
5 0

Answer: A, B and D.

Explanation:

A) If a leaf is iridescent and reflects light or colors from other objects, it helps to stand out from the rest in a low-light environment.

B) The anthocyanin pigments give the leaf a different coloration, so it has a different impact in a darker o flat colored space.

D) Having a large surface area will help to stand out from plants with small leaves.

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Air is compressed steadily from 100kPa and 20oC to 1MPa by an adiabatic compressor. If the mass flow rate of the air is 1kg/s an
igomit [66]

Answer:

(a). 575 kJ/kg.

(b). 290kw.

Explanation:

We have the following set of information or parameters from the question above;

Pressure(1) = 100kPa, Pressure (2) = 1MPa, temperature(1) = b1= 12°C = 285K = 285kJ/kg, efficiency = 80% and the mass flow rate of the air = 1kg/s.

At a temperature of 12°C, we have the value from steam table; gx= 1.2, thus gx22 = 1.2 × (1000/100) = 12.

We have that the value for b12 = 517.

Therefore, the value for h2a can be calculated as;

80/100 = (517 - 285)/ (tp at exist) - 285.

0.8 = 232/ (tp at exist) - 285.

232 = 0.8 × (tp at exist) - 285).

232 = 0.8 (tp at exist) - 228 .

(tp at exist) = 575.

Therefore, the temperature 575 kJ/kg.

Thus, the required power input of the compressor = 1kg/s × ( 575 - 285) = 290kw.

6 0
3 years ago
Savabuck University has installed standard pressure-operated flush valves on their water closets. When flushing, these valves de
Dvinal [7]

Answer:

Cost = $2527.2 per month.

Explanation:

Given that

Discharge ,Q = 130 L/min

 So

Q=0.13\ m^3/min

Cost =  $0.45 per cubic meter

1 month = 30 days

1 days = 24 hr = 24 x 60 min

1 month = 30 x 24 x 60 min

1 month = 43,200 min

Lets xm^3\ water\ waste\ in\ a\ month

x = 0.13 x 43,200

x=5616\ m^3

So the total cost = 5616 x 045 $

Cost = $2527.2 per month.

7 0
4 years ago
Poached eggs are cooked in bath of boiling water at 100°C. Over time they reach thermal equilibrium with the bath. They are then
andrew11 [14]

Answer:

a) hmax = 13334.65 w/m²k

b) t = 0.288 s

c) t = 0.693 s

d) Q = 6032.37 w = 6.032 kw

e) Transmit Heat transfer Rate into the egg is asked so it is heat of poached egg in boiling watch.

Explanation:

7 0
3 years ago
Before installing head gaskets look for “____” labels on the head gaskets
Strike441 [17]

Before installing head gaskets, you should look for "this side up" or "front" labels on the head gaskets.

<h3>What is a head gasket?</h3>

A head gasket can be defined as a gasket which is fitted between the engine block and the cylinder head in an internal combustion engine, so as to seal oil passages and absorb the pressures of the combustion that occurs inside the engine.

As a general rule, you should look for "this side up" or "front" labels on the head gaskets before installing head gaskets in an internal combustion engine of a vehicle.

Read more on head gaskets here: brainly.com/question/1264437

#SPJ1

7 0
2 years ago
The elementary liquid-phase series reaction
liraira [26]

Answer:

Concentration of A: \frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Concentration of B: \frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t } -e^{-k_{2}t } )

Concentration of C: \frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1}  } e^{-k_{1} t}

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Where x=CAo and y=CA

The mass balance for B:

-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t }-ex^{-k_{2}t }  )

The mass balance for C:

\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1}  }  e^{-k_{1}t }

The maximum concentration of C is:

C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1}  }}  =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}

and the maximum time is:

t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1}  } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h

6 0
3 years ago
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