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hram777 [196]
3 years ago
15

A 3.06kg stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What average force of air fr

iction acts on it as it falls?
Physics
1 answer:
xxTIMURxx [149]3 years ago
3 0
<span>22.5 newtons. First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as: E = 0.5mv^2 where E = Energy m = mass v = velocity Substituting known values and solving gives: E = 0.5 3.06 kg (7 m/s)^2 E = 1.53 kg 49 m^2/s^2 E = 74.97 kg*m^2/s^2 Now ignoring air resistance, how much energy should the rock have had? We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So 3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2 So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is 299.88 J - 74.97 J = 224.91 J So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division. 224.91 J / 10 m = 224.91 kg*m^2/s^2 / 10 m = 22.491 kg*m/s^2 = 22.491 N Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
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a

  v_2 =  5.6 \  m/s

b

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Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

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Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

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