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katen-ka-za [31]
3 years ago
12

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

Physics
1 answer:
garri49 [273]3 years ago
6 0
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
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Answer:

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Explanation:

From the question we are told that

       The mass of the squirrel is  m_s  =  50\ g  =  \frac{50}{1000} =  0.05 \  kg

      The surface area is   A_s =  935 cm^2  =  \frac{935}{10000} = 0.0935 \ m^2

       The height of fall is  h =4.8 m

        The length of the prism is l =  23.2 = 0.232 \ m

          The width of the prism is w =  11.6 =  0.116 \ m

 

The terminal velocity is mathematically represented as

       v_t  =  \sqrt{\frac{2 * m_s *  g }{\dho_s * C  * A } }

Where \rho  is the density of a rectangular prism with a constant values of \rho  =  1.21 \ kg/m^3

            C is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      A =  0.116 * 0.232

       A =  0.026912 \ m^2

 substituting values

      v_t  =  \sqrt{\frac{2 * 0.510 *  9.8 }{1.21 * 1  * 0.026912 } }

     v_t  =17.5 \ m/s

       

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