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3 years ago

PROBLEM IN PICTURE HELP ME DEAR GODDDDDD UGHHH NONONO I HAVE 2 MINUTES TO FINISH THIS ❕❗️❕❗️❗️❕❕❕❕❗️❕❕❗️❕❗️❗️❗️❕‼️‼️‼️‼️❗️‼️❗️

Engineering

2 answers:

Elan Coil [88]3 years ago

4 0

Thx :) so much :))))))

Julli [10]3 years ago

4 0

Answer:

Hey mate.....

Explanation:

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You might be interested in

Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper

Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

$\frac{w}{Q1} = \frac{35}{100}$

W = $1.2*\frac{35}{100}*1000kj/s$

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

$\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ$

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0

3 years ago

Oliver is designing a new children's slide to increase speed at which a child can descend.His first design involved steel becaus

Ede4ka [16]

Answer:

Steel can rusts easily in the presence of moisture and oxygen, and tarnishes as the rust progresses.

Explanation:

Steel is an alloy of carbon and iron. it is a very useful alloy that is found in almost any engineered piece. The problem with steel is that it is very susceptible to rust, and the cost of maintaining it in order to prevent rust is very high. Steel rusts in the presence of moisture and oxygen (rust is an oxidation-reduction process). Using it as a water slide exposes it constantly to moisture, and to prevent it from rusting in this case will involve a lot of maintenance cost, which is why steel is not advisable to be used in this case.

4 0

3 years ago

A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W

irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

$k_p = \dfrac{1 + sin\phi}{1 - sin\phi}$

$k_p = \dfrac{1 + sin30^0}{1 - sin30^0}$

k_p = 3

Passive earth pressure

$P_p = \dfrac{1}{2}k_p \gamma H^2$

$P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2$

P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0

3 years ago

I’m bored let’s text

Nuetrik [128]

Answer:

hi same here. hru btw

안녕하세요 모두가 잘되기를 바랍니다. 안전 유지

3 0

3 years ago

Read 2 more answers

A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60,

Rudik [331]

Answer:

for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns

Explanation:

Vp/Vs= Np/Ns

Vp: Primary voltage

Vs: Secondary Voltage

Np: number of turns on primary side

Ns: number of turns on secondary side

for output 5.6V

140/5.6= 220/Ns

Ns= 8.8 or 9 Turns

for output 12.0V

140/12= 220/Ns

Ns= 18.9 or 19 turns

for output 480V

140/480= 220/Ns

Ns= 754.3 or 755 turns

4 0

4 years ago