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3 years ago

What are the 5 major forest types?

Engineering

2 answers:

Misha Larkins [42]3 years ago

8 0

Balls balls baldbsisvsbsbsnnsjshhshsjsjjsjdjdndj j j j I j g f h y f j j n me. K I DVD’s by

Nataly [62]3 years ago

4 0

Answer:

1. Equatorial Evergreen or Rainforest

2. Tropical forest

3. Mediterranean forest

4. Temperate broad-leaved forest

5. Warm temperate forest

Explanation:

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Which option is an example of a physical model?

Burka [1]

Answer:

A. A clay ball with a slice cut out, showing the layers of the earth

Explanation:

6 0

3 years ago

How do you get A's in school

soldier1979 [14.2K]

By being attentive in classes, taking good notes and making sure to put time fourth into your grades.

5 0

2 years ago

Read 2 more answers

The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f

kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

$C_{v} =AT^{3}$

Substituting in the expression of the question:

$C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3} } )T^{3}$

$\theta _{D} =(\frac{12\pi ^{4}RT^{3} }{5C_{v} } ) ^{1/3}$

Here

$C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K$

Replacing:

$\theta _{D} =(\frac{12\pi ^{4}*8.31*15^{3} }{5*0.1241} )^{1/3} =375.2361K$

3 0

2 years ago

A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium

Juli2301 [7.4K]

Answer:

N_A=1.5*10^-8 kmol/s.m^2

Explanation:

KNOWN:

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.

FIND:

Molar diffusion flux.

ASSUMPTIONS:

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.

ANALYSIS: The molar flux may be obtained from

N_A=D_AB/L(C_A,1-C_A,2)

=10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3

N_A=1.5*10^-8 kmol/s.m^2

COMMENTS: The mass flux is:

n_A,x=M_a*N_A,x

n_A,x=6*10^-8 kg/s m^2

5 0

2 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The vol

SCORPION-xisa [38]

Answer:

heat transfer rate is -15.71 kW

Explanation:

given data

Initial pressure = 4 bar

Final pressure = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1 = m2

and mass flow equation is express as

m1 = $\frac{A1\times V1}{v1}$ .......................1

m1 = $\frac{4\times \frac{1}{60}}{0.05397}$

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + $\frac{v1^2-v2^2}{2}$ + g (z1-z2) ] ....................2

so here Qcv will be

Qcv = m × [ $\frac{Wcv}{m} + (h2-h1)$ ] ......................3

put here value and we get

Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv = -15.7130 kW

so here heat transfer rate is -15.71 kW

6 0

2 years ago