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Jobisdone [24]
3 years ago
5

Without Newton’s third law of motion, what would an object sitting on a table do?

Physics
2 answers:
Angelina_Jolie [31]3 years ago
8 0

Newton's third law: If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

boyakko [2]3 years ago
6 0

Without Newton's third law of motion, an object sitting on a table would continue sitting there, motionless, just we always see it do, and just as objects used to do before Newton ever developed his laws of motion.  

Getting the new law didn't change anything.  Just like everything else in Science, the new law was just a new description of how Nature works.  But it didn't change anything, because Nature doesn't care whether we understand how it works or not.

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A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
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Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

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

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

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