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Vladimir [108]
3 years ago
6

While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba

ckstretch. To train for the 800m event, he runs both the 100m homestretch and the 100m backstretch in 12.2 seconds. How do Milt's velocities on the homestretch and backstretch compare? A) The homestretch and backstretch have velocities with the same magnitude only. B) The backstretch velocity is greater in magnitude than the homestretch velocity. C) The home stretch velocity is greater in magnitude than the backstretch velocity. D) The homestretch and the backstretch have velocities with the same direction and magnitude.
Physics
2 answers:
Yuri [45]3 years ago
8 0
Answer a would be correct since velocity is a vector and has a magnitude and a direction. In this case v₁ = - v₂.
Wittaler [7]3 years ago
5 0

The correct answer is

A) The homestretch and backstretch have velocities with the same magnitude only.

Explanation:

Velocity is a vector - it means it consists of:

1) A magnitude: called speed, it is given by the ratio between the total distance covered, d, and the time taken, t:

v=\frac{d}{t}

2) A direction: the direction of motion

In this example, we see that the homestretch and the backstretch have same speed, because the same distance (100 m) is covered in the same time interval (100 m), however the direction is different (the homestretch is due East, while the backstretch is due West). So, the correct answer is

A) The homestretch and backstretch have velocities with the same magnitude only.

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Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

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          ∫ (α x³ + β) dx = ΔK

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Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
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Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
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