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3 years ago

What is the velocity in m/s of a rocket that goes 461..6 km north in 44.7 seconds?

Physics

2 answers:

Jlenok [28]3 years ago

6 0

Velocity = (displacement) / (time)

Velocity = (change in location) / (time for the change), and direction of the change

Displacement = (location at the end) minus (location at the start)

Displacement = (461.6 km North) - (zero) = 461.6 km North

Velocity = (461.6 km North) / (44.7 sec)

Velocity = (461.6 / 44.7) km/sec North

Velocity = 10.33 km/sec North

That's right about 23,100 miles per hour ! That rocket is really honkin !

With just a little more oomph, that rocket could go up and NEVER come back. The 'escape velocity' from Earth is 11.2 km/sec, and this rocket already has 92% of it !

bekas [8.4K]3 years ago

4 0

10.32 km/s

(V = distance/time)

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From Newton's second Law of Motion

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7nadin3 [17]

Answer:

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(b) r = $\frac{33}{25} R_E$

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

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$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

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(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

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