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melisa1 [442]
3 years ago
10

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Physics
1 answer:
Sidana [21]3 years ago
7 0

Answer:

acceleration of the rocket is given as

a = 7.45 m/s^2

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

v_i = 0

v_f = 447 m/s

t = 1 min = 60 s

so we have

a = \frac{v_f - v_i}{t}

a = \frac{447 - 0}{60}

a = 7.45 m/s^2

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DE which is the differential equation represents the LRC series circuit where
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3 years ago
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Problems - Show all work.
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