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Leviafan [203]
3 years ago
9

The force on a bullet that results from summing all of the forces on that bullet is known as the _____ on the bullet.

Physics
2 answers:
statuscvo [17]3 years ago
4 0

Answer:

The force on a bullet that results from summing all of the forces on that bullet is known as the net force on the bullet.

Explanation:

Net force represents the overall force or the total force acting on an object.

Here, the object is bullet and the net force acting on it is the addition of all the forces acting on the bullet.

NeX [460]3 years ago
3 0
Net force.

Check and see if this is included in the vocabulary for this subject, there are a few different ways to say this, but this is the most popular (and the one I am most familiar with).
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A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
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Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

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}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

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

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

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