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2 years ago

The National Park Service is waiting on your recommendation.

Engineering

1 answer:

maksim [4K]2 years ago

8 0

The hypothesis that would help restore equilibrium is reintroduce wolves in the park or into the forest.

<h3>What is reintroduce?</h3>

To reintroduce, is a term that means to revive, reinstate, or bring back something.

Note that based on the right recommendation needed by National Park Service, one can say that the best hypothesis that would help restore equilibrium is reintroduce wolves in the park or into the forest as it will help restore balance.

Learn more about hypothesis from

brainly.com/question/24149728

#SPJ1

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A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition Cβ = 87

we use here lever rule that is

Wα = Wβ ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0

4 years ago

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost m

Ilia_Sergeevich [38]

Answer:

ordinary bulb total cost is $39.54

fluorescent bulb total cost is $13.05

amount save = 39.54 - 13.05 = $26.49

resistance = 626.1 ohm

Explanation:

in the 1st part

bulb on time = 3 year = 4380 hours

life of bulb = 750 h

so number of bulb required = $\frac{4380}{750}$

number of bulb required = 6

cost of 6 bulb is = 6 × 0.75 = $4.5

cost of operation is = 100 × 4380 × $\frac{0.08}{1000}$

cost of operation = $35.04

so total cost will be = $4.5 + $35.04 = $39.54

and

when compare with florescent bulb

time = 3 year = 4380 h

life of bulb = 10000 h

so number of bulb required = $\frac{4380}{10000}$

number of bulb required = 0.43 = 1

cost of 6 bulb is = 1 × 5 = $5

cost of operation is = 23 × 4380 × $\frac{0.08}{1000}$

cost of operation = $8.05

so total cost will be = $5 + $8.05 = $13.05

in part 2nd

total amount save while compare bulb is

amount save = 39.54 - 13.05 = $26.49

and in part 3rd

resistance of bulb is

resistance = $\frac{v^2}{P}$

resistance = $\frac{120^2}{23}$

resistance = 626.1 ohm

6 0

3 years ago

Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and

ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, $Q=16 m^{3}/min$

Diameter d= 8cm= 0.08 m

Assumptions

The flow is jet flow hence momentum-flux correction factor is unity
Gravitational force is not considered
The flow is steady, frictionless and incompressible
Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

$V=\frac {Q}{A}$ where V is the exit velocity and A is area

Area, $A=\frac {\pi d^{2}{4}$ where d is the diameter

By substitution

$V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min$

To convert v to m/s from m/s, we simply divide it by 60 hence

$V=\frac {3183.098862 m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s$

3 0

3 years ago

Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate h

shusha [124]

Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate himself for the Waterman Award at the National Science Foundation. His earlier emails were offered where he had solicited letters were offered as evidence. Siddiqui claimed that content of earlier emails was hearsay. Do the earlier emails come in is given below

Explanation:

1.Mohamed Siddiqui appeals his convictions for fraud and false statements to a federal agency, and obstruction in connection with a federal investigation. Siddiqui challenges the district court's admission into evidence of e-mail and foreign depositions.

2.On February 18, 1997, Jodi Saltzman, a special agent with the NSF interviewed Siddiqui at Siddiqui's office at the University of South Alabama. During the interview, Siddiqui signed a statement admitting that he had nominated himself for the Waterman Award, but that he had permission from Yamada and von Gunten to submit forms on their behalf. Siddiqui also acknowledged in the statement that Westrick had recommended Siddiqui for a different award, the PECASE Award, but that Siddiqui had changed the wording of the letter to apply to the Waterman Award. Siddiqui was indicted on April 29, 1997.

3.Siddiqui opposed the taking of the depositions on the grounds that the witnesses' personal presence at trial was necessary, and that Indian travel restrictions for its citizens residing abroad prevented him from traveling to Japan and Switzerland. Specifically, Siddiqui asserted that because of religious persecution in India his travel to Japan or Switzerland related to the criminal action would put his family members still living in India at risk. The magistrate judge ruled that the government had carried its burden of showing that Yamada and von Gunten would be unavailable to appear at trial, and instructed that Siddiqui's fear of obtaining a travel visa from India because of the threat of persecution of family members should not preclude the taking of the foreign depositions.

4.Yamada's deposition was taken in Japan on March 6, 1998. At government expense, Siddiqui's counsel attended the deposition and cross-examined the witness, but was not in telephonic contact with Siddiqui during the deposition. Yamada testified that on February 1, 1997, she received an e-mail stating that if she received a phone call from the NSF to “please tell good words about me.” Yamada testified that she knew the e-mail was from Siddiqui because the name on the e-mail had Siddiqui's sender address, and it ended with the name “Mo” which Siddiqui had previously told her was his nickname, and which he had used in previous e-mail.

5.Yamada later admitted to Saltzman that she had not given Siddiqui permission to sign, but had made the earlier representation because she thought Siddiqui would go to jail.

6.During cross-examination of Yamada at the deposition, Siddiqui's counsel introduced an e-mail from Yamada to Siddiqui. This e-mail contained the same e-mail address for Siddiqui as the e-mail received by Yamada and von Gunten apparently from Siddiqui.

7.Von Gunten's video deposition was taken in Switzerland. At government expense, Siddiqui's counsel attended the deposition and cross-examined von Gunten. During the deposition, Siddiqui was in communication with his counsel by telephone. Von Gunten testified at the deposition that he had not submitted a letter of recommendation in favor of Siddiqui for the Waterman Award, and that he had not given Siddiqui permission to submit such a letter in his name.

8 0

4 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0

3 years ago