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Ilia_Sergeevich [38]
3 years ago
9

What material is rare in the atmosphere of earth, but common in the atmospheres of mars and venus?

Physics
1 answer:
Oksanka [162]3 years ago
4 0
The answer is carbon dioxide
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Captain Hook is ghting Peter Pan, and they are about to step onto a tightrope strung horizontally between two maststhat are 16 m
andreyandreev [35.5K]

Complete Question

Captain Hook is fighting  Peter Pan, and they are about to step onto a tightrope strung horizontally between two masts that are 16 m apart. When Pan and Hook are standing exactly halfway between the masts, the rope makes a 3 anglewith the horizontal. The rope has a diameter of 0.02 m and a Youngs modulus of 35 GPa.

What is the combined mass,in kilograms, of Peter Pan and Captain Hook?

Answer:

Their combined mass is m= 161.2kg

Explanation:

A sketch that describes the question is shown on the first uploaded image  

  From the question we are told that

          The distance apart is d_A = 16m

          The angle the rope makes is \theta = 3^o

          The diameter of the rope is d = 0.02m

          The Young modulus is  Y = 35Pa

From the diagram we see that the elongation of the rope can be  mathematically evaluated as

         \Delta L = 2x - 16

And applying  SOHCATOH rule    x = \frac{8}{cos \theta}

   Substituting values

                              x = \frac{8}{cos (3)}

                                = 8.01m

      And   \Delta L = \frac{16}{cos 3}  -16

                     \Delta L = 0.02196m

The Tension on the rope can be mathematically represented as

               T = Y A * \frac{\Delta L}{L}

Where A is the area and is mathematically represented as

              A = \frac{\pi}{4} d^2

 Substituting values

            A = \frac{\pi}{4} (0.02)^2

Now Substituting values into the formula for the tension on the rope

          T = (35*10^9) * \frac{\pi}{4} (0.02)^2 * \frac{(0.02196)}{16}

             =15093.4 N

From the diagram we can mathematically evaluate the the weight of peter and hook as

              W = 2T sin \theta

Where W = mg

Now substituting this into the equation and making m the subject

                   m = \frac{2Tsin \theta}{g}

Substituting values

                m = \frac{2* 15093.4 sin(3)}{9.8}

                    m= 161.2kg

Note  SOHCATOH is

                         Sin \theta = \frac{opposite}{hypotenuse}\\ Cos \theta = \frac{adjacent }{hypotenuse} \\Tan \theta = \frac{opposite}{adjacent}      

8 0
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