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3 years ago

Name some of the traits that Darwin studied in the Galapagos finches.

Engineering

2 answers:

Anna11 [10]3 years ago

8 0

Answer:

Darwin noticed that fruit-eating finches had parrot-like beaks, and that finches that ate insects had narrow, prying beaks. He wrote: "One might really fancy that from an original paucity [scarcity] of birds ... one species had been taken and modified for different ends."

yKpoI14uk [10]3 years ago

7 0

What grade is this in?

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Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)

vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0

3 years ago

All PPE used for lung protection must be tight-fitting. Is this statement TRUE or FALSE?

lubasha [3.4K]

I think true I’m not sure

3 0

2 years ago

An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener

RSB [31]

Given:

$I_{rms} = 50 A$

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, $R = 100\ohm$

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = $50\times 3600$ W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = $\frac{P}{V'}$

I = $\frac{1800000}{100000}$

I = 1.8 A

Power line voltage drop is given by:

$V_{drop} = I\times R$

$V_{drop} = 1.8\times 100$

$V_{drop} = 180 V$

Power dissipated in long transmission line $P_{dissipated} = V_{drop}\times I$

Power dissipated in long transmission line $P_{dissipated} = 180\times 1.8$ = 324 W

% Heat loss in power line, $P_{loss} = \frac{P_{dissipated}}{P}\times 100$

% Heat loss in power line, $P_{loss} = \frac{324}{180000}\times 100$

$P_{loss} = 0.18%$

5 0

3 years ago

Noooo plssschvwekjshdjkshdjkshdjksahdk

sashaice [31]

Huh? Kffjkdkdodjenxkrkdkre

4 0

3 years ago

Read 2 more answers

A logic chip used in a computer dissipates 3 W of power in an environment at 120°F, and has a heat transfer surface area of 0.08

UkoKoshka [18]

Answer:

attached below

Explanation:

7 0

4 years ago