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alexira [117]
3 years ago
9

An object of mass 0.50 kg is transported to the surface of Planet X where the object’s weight is measured to be 20 N. The radius

of the planet is 4.0 x 10^6 m. What free fall acceleration will the 0.50-kg object experience when transported to a distance of 2.0 x 10^6 m from the surface of this planet?
Physics
2 answers:
Olin [163]3 years ago
8 0

Answer:

g'=40\ m.s^{-2}

Explanation:

Given:

  • mass of the object, m=0.5\ kg
  • weight of the object on planet x, W=20\ N
  • radius of the planet, R=4\times10^6\ m
  • radial distance between the planet and the object, r=2\times 10^6\ m

<u>Now free fall acceleration on planet X:</u>

W=m.g'

20=0.5\times g'

g'=40\ m.s^{-2} irrespective of the height.

sveticcg [70]3 years ago
3 0

Answer:

17.78 m/s^2

Explanation:

m = 0.5 kg

Weight on the planet, W = 20 N

Acceleration due to gravity on the surface of planet, g = W / m

g = 20 / 0.5 = 40 m/s^2

Radius of planet, r = 4 x 10^6 m

height, h = 2 x 10^6 m

Let the acceleration due to gravity at a height is g'.

According to the formula of acceleration due to gravity at height is

g' = g\frac{R^{2}}{(R+h)^{2}}

g' = 40\frac{\left (4\times 10^6  \right )^{2}}{(4+2)^{2}\times 10^{12}}

g' = 17.78 m/s^2

Thus, the acceleration due to gravity at height is  17.78 m/s^2 .

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
photoshop1234 [79]

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

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