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Natali5045456 [20]
3 years ago
9

Determine how many gallons of jet fuel the jet can hold and still be able to takeoff from sea level with an outside temperature

of 277 K. The specific gravity of jet fuel is 0.804.
Physics
1 answer:
zysi [14]3 years ago
4 0

Question:

For a certain commercial jet to be able to achieve liftoff, its total weight cannot exceed 333,390 kg. If the empty weight (i.e., no fuel or passengers) of the jet is 163,293 kg and there are 300 passengers aboard with an average weight of 75.0 kg, determine how many gallons of jet fuel the jet can hold and still be able to takeoff from sea level with an outside temperature of 277 K. The specific gravity of jet fuel is 0.804

Answer:

The answer to the question is;

The jet can hold 48642.2 gal or 40503 galUK of jet fuel and still be able to takeoff from sea level.

Explanation:

To solve the question, we note the following

Maximum total weight at liftoff = 333,390 kg

Mass of jet when empty = 163,293 kg

Average mass of one passenger = 75.0 kg

Number of persengers = 300

Total mass of persengers = 300×75 =22500 kg

Mass of  jet + passengers = 163293 + 22500 = 185793 kg

Allowable mass jet carry = Mass of jet fuel

= Maximum total weight at liftoff - Mass of  jet + passengers  

= 333,390 kg - 185,793 kg = 147597 kg

Specific gravity of jet fuel = 0.804 therefore density of jet  fuel  is given by

Density of jet fuel = Density of water × The specific gravity of jet fuel

= 997 kg/m³ × 0.804 = 801.588 kg/m³

Volume of jet fuel = (Mass /Density) of the jet fuel = \frac{147597 }{801.588} = 184.131 m³

However 1 m³ = 264.172 gal  = 219.969 galUK

∴ 184.131 m³ = 264.172× 184.131  gal

= 48642.2 gal = 40503 galUK

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3 years ago
A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns
Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

f'=\frac{V+V_{o}}{V-V_{s}} f  (1)

Where:

f=250,000 Hz is the actual frequency of the sound wave

f'=240,000 Hz is the "observed" frequency

V=340 m/s is the velocity of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which is the balloon

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}  (2)

V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}  (3)

V_{s}=-14.16 m/s (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (t=4.8 s) given as data, we can find the distance:

d=V_{s}t (5)

d=(14.16 m/s)(4.8 s) (6)

Finally:

d=68 m (8) This is the distance of the balloon from the radar station

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