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FrozenT [24]
3 years ago
15

Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .

Physics
1 answer:
Radda [10]3 years ago
7 0

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{160000}{270000}  =  \frac{16}{27}  \\  = 0.592592...

We have the final answer as

<h3>0.59 m/s²</h3>

Hope this helps you

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T_b=107.3784\ ^{\circ}C

Explanation:

Given:

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  • radius of the base of the kettle, r=0.12\ m
  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
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<u>So, the heat rate:</u>

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<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

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A= area of the surface through which conduction occurs

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15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

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