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1 year ago

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The fuel for a jet engine has a density of 1.32 slug/ft3. if the total volume of fuel tank a is 50 ft3, determine the weight of

the fuel when the tanks are completely full?

1 answer:

inessss [21]1 year ago

6 0

The weight of the fuel when the tank is filled is 2.13 kip.

The density of the fuel of the jet engine= 1.32 slug/ft³

The total volume of the fuel that can be filled in the fuel tank of a jet engine = 50 ft³

The specific weight of the fuel when the tank is filled is,

$\gamma = p \times g$

$\gamma = 1.32 \times 32.2$

$\gamma = 42.504 \: ib/ft ^{3}$

The specific weight of the fuel when the tank is filled is 42.504 ib/ft ³.

The weight of the fuel when the tank is filled is,

$w = \gamma \times ∀$

$w = 42.5.4 \times 50$

$w = 2.13 \: kip$

The weight of the fuel is 2.13 kip.

Therefore, the weight of the fuel when the tank is filled is 2.13 kip.

To know more about specific weight, refer to the below link:

brainly.com/question/13178675

#SPJ4

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

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Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

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Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

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