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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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A cruise ship is having troubles with buoyancy. What is a reasonable solution? A. Increase the weight of the ship above water B.
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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul
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Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

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ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

Now, this force is no other than centripetal force, so we can write this equation:

Ff  = Fc ⇒ μ*m*g = m*ac

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c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed  relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

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a = √(ac)²+(at)² = 0.67 m/s²

Now, as friction force opposes to the relative movement between surfaces (the seed and the turntable), it shall be larger than the product of the mass times the total acceleration, acting along  the same action line, so we can say:

Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

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