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3 years ago

14

Which of the following might be a job or task of an IT worker who manages

1 answer:

-Dominant- [34]3 years ago

7 0

Answer:

I'm pretty sure it's letter d

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help me please help me please help me please help me please help me please help me please help me please help me please help me

kakasveta [241]

Answer:

With what?

Explanation:

Brainliest?

7 0

3 years ago

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The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th

Alja [10]

Answer:

Explanation:

$y'''+y=0---(i)$

General solution

$y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)$

$y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1$

in equation (iii)

$c_1=c_2=-1$

Therefore,

$\large\boxed{y=-1-\cos x-2\sin x}$

5 0

3 years ago

Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl

Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

$A_1$ = 1 m²

$P_1$ = 100 kPa

$V_1$ = 180 m/s

Flow rate

$Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s$

Volumetric flow rate = 180 m³/s

Mass flow rate

$\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s$

Mass flow rate = 213.4 kg/s

3 0

3 years ago

To read signs you need good focal vision

kow [346]

Answer:eyesight

Explanation:

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3 years ago

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At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

tangare [24]

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

Diameter of the turbine = 90 m

Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

$E=\dfrac{1}{2}mv^2\ J$

That is why mechanical energy per unit mass will be

$e=\dfrac{1}{2}v^2\ J/kg$

Now by putting the values in the above equation we get

$e=\dfrac{1}{2}\times 10^2\ J/kg$

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

5 0

3 years ago