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Naily [24]
3 years ago
10

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

surements: a 2.50-kg stone thrown upwardfrom the ground at 15.0 {\rm m/s} returns to the ground in 9.00 {\rm s}; the circumference of Mongo at the equator is1.00×105 {\rm km} ; and there is no appreciable atmosphere onMongo.part AThe starship commander, Captain Confusion,asks for the following information: what is the mass ofMongo?m ={\rm kg}part bIf the Aimless Wanderer goes into acircular orbit 2.00×104 {\rm km} above the surface of Mongo, how many hours will ittake the ship to complete one orbit?t ={\rm h}
Physics
1 answer:
Setler [38]3 years ago
3 0

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

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Answer:

0.2012 m/s

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Explanation:

M = mass of ice hockey goalie = 70 kg

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Using conservation of momentum

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7.055 = (70) V' + (0.170) v'

V' = 0.1008 - 0.00243 v'                                       eq-1

Using conservation of kinetic energy

(0.5) M V² + (0.5) m v² =  (0.5) M V'² + (0.5) m v'²

M V² + m v² = M V'² + m v'²

(70) (0)² + (0.170) (41.5)² = (70) V'² + (0.170) v'²

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292.8 = (70) (0.1008 + 0.00243 v')² + (0.170) v'²

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Using eq-1

V' = 0.1008 - 0.00243 v'

V' = 0.1008 - 0.00243 (- 41.3)

V' = 0.2012 m/s

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