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Rudik [331]
3 years ago
5

In 1865, jules verne proposed sending men to the moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

97 km/s. what would have been the unrealistically large acceleration experienced by the space travelers during their launch? (a human can stand an acceleration of 15g for a short time.)
Physics
1 answer:
Aneli [31]3 years ago
5 0
I'm going to assume that the 10.97 km/s is the velocity at the end of the cannon.

vi = 0
d = 220m
vf = 10.97 km/s = 10970 m/s
a = ???

vf^2 = vi^2 + 2*a*d
10970^2 = 0 + 2*a*220
10970^2 / 440 = a You are right this is a pretty big number.
2.73 * 10^5 = a


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Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

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   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

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A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mas
Sergio039 [100]

Complete Question

The diagram for this question is shown on the first uploaded image

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The distance traveled in horizontal direction is D = 1.38 m

Explanation:

From the question we are told that

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Generally the work energy theorem can be mathematically represented as

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