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lana66690
3 months ago
5

A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times

the mass of the empty rocket. Part A What is the rocket's speed when half the fuel has been burned?
Physics
1 answer:
Sav [3.1K]3 months ago
3 0

Answer:

 v_{f} = 1,386 m / s

Explanation:

The mechanism behind rocket propulsion is defined by the formula

       v_{f} - v₀ =  v_{r} ln (M₀ / Mf)

Here, v refers to the initial, final, and relative velocities, while M indicates the masses

The provided values include the relative velocity (see = 2000 m / s) and the initial mass, where the mass of the rocket when loaded is represented as (M₀ = 5Mf)

For our analysis, we assume the rocket begins at rest (v₀ = 0)

Once half of the fuel has burnt, the mass ratio indicates that the current mass is    

       M = 2.5 Mf

       v_{f} - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

       v_{f} = 1,386 m / s

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In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
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Answer:

The period for the first satellites is 24.46 days, while the second satellites have a period of 37.67 days

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\frac{T}{r^{\frac{3}{2} } } = constant

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\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For the second satellites,

\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Thus, the orbital period for the first satellites is 24.46 days and for the second satellites, it is 37.67 days

8 0
2 months ago
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