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Vladimir79 [104]
3 years ago
15

An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only

a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion, away from the space station. The astronaut has a mass of 124 kg and the bag of tools has a mass of 16.0 kg. If the astronaut is moving away from the space station at ????i=1.80 m/s initially, what is the minimum final speed of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?
Physics
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

15.75 m/s

Explanation:

v = Velocity of the combined mass of astronaut and tools = 1.8 m/s

m_1 = Mass of astronaut = 124 kg

m_2 = Mass of tools = 16 kg

v_1 = Velocity of astronaut = 0

v_2 = Velocity of tools

As linear momentum is conserved

m_1v_1 + m_2v_2 =(m_1 + m_2)v\\\Rightarrow v_2=\frac{(m_1 + m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(124+16)\times 1.8-124\times 0}{16}\\\Rightarrow v_2=15.75\ m/s

The velocity of the tools is 15.75 m/s

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Answer:

12.0 m/s²

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a = F/m

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a' = ⅔ (F/m)

a' = ⅔ (18.0 m/s²)

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A yellow train of mass 100 kg is moving at 8 m/s towards an orange train of mass 200 kg traveling on the opposite direction on t
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To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which

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Since the question only wants the sum of initial momentum,

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8 0
2 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
2 years ago
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