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yawa3891 [41]
3 years ago
9

5. If you walk 70 meters north in 5 minutes and 20 meters south in 15 seconds. What was the average

Physics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

kjfvkuzfvnv

Explanation:

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A force of 5 N accelerates an object. The object’s mass is 50 g. What is the acceleration of the object? (Formula: F=ma)
natima [27]

Explanation:

F = ma

5 N = (0.050 kg) a

a = 100 m/s²

3 0
3 years ago
The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du
leonid [27]

Answer:

The magnitude of the force is  F_{net}= 1.837 *10^4N

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

      At t = 0 , \theta = 20^o

      The rate at which the angle increases is w = 2 \ ^o/s

Converting this to revolution per second  \theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps

     The length  of the rope is defined by

                      r = 125- \frac{1}{3}t^{\frac{3}{2} }    

    At \theta  =30^o , The tension on the rope T = 18 kN

      Mass of the para-sailor is M_p = 75kg

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

  Now the derivative of displacement is velocity

   So

           r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }

represents the velocity, again the derivative of velocity gives us acceleration

So

         r'' = -\frac{1}{4} t^{-\frac{1}{2} }

Now to the time when the rope made angle of 30° with the water

      generally angular velocity is mathematically represented as

                      w = \frac{\Delta \theta}{\Delta t}

Where \theta is the angular displacement

      Now considering the interval between 20^o \ to \ 30^o we have

                 2 = \frac{30 -20 }{t -0}

making t the subject

             t = \frac{10}{2}

               = 5s

Now at this time the displacement is

             r = 125- \frac{1}{3}(5)^{\frac{3}{2} }  

                = 121.273 m

The linear velocity is

             r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }

                = -1.118 m/s

The linear acceleration is

          r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }

              = -0.112m/s^2

Generally radial acceleration is mathematically represented by

         \alpha _R = r'' -r \theta'^2

              = -0.112 - (121.273)[0.0349]^2

              = 0.271 m/s^2

Generally angular acceleration  is mathematically represented by

                 \alpha_t = r \theta'' + 2 r' \theta '

Now \theta '' = \frac{d (0.0349)}{dt}  = 0

So

             \alpha _t = 121.273 * 0  + 2 * (-1.118)(0.0349)

                   = -0.07805 m/s^2

The net resultant  acceleration is mathematically represented as

                a = \sqrt{\alpha_R^2 + \alpha_t^2  }

                  = \sqrt{(-0.07805)^2  +(-0.027)^2}

                  = 0.272 m/s^2

Now the direction of the is acceleration is mathematically represented as

                  tan \theta_a = \frac{\alpha_R }{\alpha_t }

                       \theta_a = tan^{-1} \frac{-0.271}{-0.07805}

                           = 73.26^o

               

The force on the para-sailor along y-axis is mathematically represented as

               F_y = mg + Tsin 30^o + ma sin(90- \theta )

                    = (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)

                    = 9.74*10^3 N

The force on the para-sailor along x-axis is mathematically represented as

              F_x = mg + Tcos 30^o + ma cos(90- \theta )    

             = (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)

             = 1.557 *10^4 N

The net resultant force is mathematically evaluated as

                      F_{net} = \sqrt{F_x^2 + F_y^2}

                             =\sqrt{(1.557 *10^4)^2  + (9.74*10^3)^2}

                            F_{net}= 1.837 *10^4N

The direction of the force is

              tan \theta_f = \frac{F_y}{F_x}

                   \theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]

                       = tan^{-1} (1.599)

                       = 57.98^o

     

                     

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