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3 years ago

If an imbalance occurs, the _

Engineering

1 answer:

pochemuha3 years ago

7 0

A. AFGI is the answer for this question.

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ممكن الحل ............

Roman55 [17]

Answer:

i dont understand

Explanation:

4 0

3 years ago

A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much

Andrej [43]

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

4 0

3 years ago

In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o

lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

= $10\times 101325 \ Pa$

= $1013250 \ Pa$

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

= $1 \ m^3$

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ $PV=nRT$

⇒ $n=\frac{PV}{RT}$

By substituting the values, we get

$=\frac{1013250\times 1}{8.3145\times 298}$

$=408.94 \ moles$

As we know,

⇒ $Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}$

or,

⇒ $MW=\frac{m}{n}$

$=\frac{11.5}{408.94}$

$=0.02812 \ Kg/mol$

$=28.12 \ g/mol$

3 0

3 years ago

Rear axles are usually lubricated by the same gear oil that lubricates the differential. True or false

KIM [24]

Answer:

true

Explanation:

8 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