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andrew11 [14]
2 years ago
8

A 1500 kg aircraft going 35 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and

are going 15 m/s after the collision. If they skid for 112.1 m before stopping, how long (in seconds) did they skid? Hint: Are the aircraft moving at a constant velocity after the collision or do they experience an acceleration?
Physics
1 answer:
Elis [28]2 years ago
3 0

The two aircrafts are skidded for 14.9 s.

To find the answer, we need to know about the Newton's equation of motion.

<h3>What is the Newton's equation that relates velocity, distance, acceleration and time?</h3>

As per Newton's equation of motion

  • V²-U²= 2aS
  • V= U+at
  • V= final velocity, U = initial velocity, S = distance, a= acceleration, t= time

<h3>What's the acceleration of the aircrafts that skidded from 35 m/s after the collision for 112.1 m before achieving 15 m/s?</h3>
  • Here, U = 35 m/s, V = 15m/s, S= 112.1 m
  • So, 15²- 35²= 2a×112.1= 224.2a

=> a= -1000/224.2= -4.5 m/s²

<h3>What's the time taken to stop if these aircrafts are de-accelerated by 4.5 m/s² from 35m/s?</h3>
  • Here, U = 35 m/s, V=15 m/s a= -4.5 m/s²
  • 35= 15+(-4.5)t

=> t= 20/4.5 = 4.4 s

Thus, we can conclude that the two aircrafts are skidded for 4.4 s.

Learn more about the Newton's equation of motion here:

brainly.com/question/25545050

#SPJ1

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7 0
3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
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Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

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          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

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       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

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Explanation:

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minimum wavelength =  2.78 m

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