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2 years ago

Part D

Engineering

2 answers:

Tatiana [17]2 years ago

8 0

Answer:

clothes iron its 3rdyeyeyey

Explanation:

clothes iron its 3rd

btw quaotra

yesss

victus00 [196]2 years ago

5 0

Answer:

I would say 185

Explanation:

If a TV is 150 watts and a video game sistem is 35 then you would have to add them together and you would then get 185. So a TV connected to a video game system would run at 185 watts.

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The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago

A clean machine is a _______________ machine.

solniwko [45]

A clean machine is a clean machine :-)

4 0

2 years ago

A cyclic tensile load ranging from 0 kN to 55 kN force is applied along the length of a 100 mm long bar with a 15 mm x 15 mm squ

Yuliya22 [10]

Answer:

square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.

Required:

a. What is the nominal maximum tensile stress on the bar?

b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?

7 0

3 years ago

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

2 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago