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7nadin3 [17]
3 years ago
8

The fastest man alive can run the 100 meter dash in 9.58 seconds, calculate average speed in meters per second

Physics
1 answer:
densk [106]3 years ago
3 0

Answer:

100 ÷ 9.58 = 10.44 (approximate answer)

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Zoning laws establish _______. a. what types of buildings can be built in an area b. the uses an area of land can be put to c. w
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A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and
Alexxandr [17]

Answer:

a

  F  =326.7 \ N

b

  M  = 6

Explanation:

From the question we are told that

          The mass of the rock is  m_r  =  200 \ kg

          The  length of the small object from the rock is  d  =  2 \ m

          The  length of the small object from the branch l  =  12 \ m

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      W =  m_r *  g

substituting values

     W =   200 *  9.8

     W =   1960 \ N

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         \sum  M_f  =  F * cos \theta *  l  -  W cos\theta  *  d =  0

Here  \theta is very small so  cos\theta  *  l  =  l

                               and  cos\theta  *  d  =  d

Hence

       F *   l  -  W  * d =  0

=>    F  = \frac{W * d}{l}

substituting values

        F  = \frac{1960 *  2}{12}

       F  =326.7 \ N

The  mechanical advantage is mathematically evaluated as

          M  = \frac{W}{F}

substituting values

        M  = \frac{1960}{326.7}

       M  = 6

6 0
3 years ago
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