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devlian [24]
3 years ago
11

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

e, the ship's orbital speed is 5500 m/s . By observing the planet, you determine its radius to be 4.48×106 m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 14.6 m/s at an angle of 30.8∘ above the horizontal.If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile?
Physics
1 answer:
Anarel [89]3 years ago
3 0

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

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Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t
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According to newton's law of gravitation, the gravitational force(F)  is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

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Combining equation (1) and (2),

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This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

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8 months ago
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
Jobisdone [24]

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
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 35 m/s. The ball has __________________
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Answer:

The ball has kinetic energy

the kinetic energy is 945 J

Explanation:

4 0
2 years ago
Read 2 more answers
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