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ankoles [38]
3 years ago
8

This procedure has been used to "weigh" astronauts in space: A 42.5 kg chair is attached to a spring and allowed to oscillate. W

hen it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair now takes 2.54 s for one cycle.What is the mass of the astronaut?
Physics
1 answer:
Bingel [31]3 years ago
3 0

Answer:

The mass of the astronaut is approximately 119.74 kg

Explanation:

Assuming this problem as a Simple Harmonic Motion of a mass-spring system, the period (T) of the oscillations for a mass (m) and spring constant (k) is:

T=2\pi\sqrt{\frac{m}{k}} (1)

First, we have to calculate the spring constant using equation (1) and the data provided for the oscillations without the astronaut:

<em>(it’s important to note that one complete vibration is the period of the movement)</em>

1.3=2\pi\sqrt{\frac{42.5}{k}}\Longrightarrow k=42.5(\frac{2\pi}{1.3})^{2}

k\approx992.8\,\frac{N}{m}

Now with the value of k, we can use again (1) to find the mass of the astronaut (Ma) that makes the period to be 2.54 seconds

2.54=2\pi\sqrt{\frac{42.5+M_{a}}{992.8}}\Longrightarrow M_{a}=992.8(\frac{2.54}{2\pi})^{2}-42.5

\mathbf{M_{a}\approx119.74\,kg}

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Answer:

a = 4.9(1 - sinθ - 0.4cosθ)

Explanation:

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

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     The angle of inclination of the second block is  \theta _2 =  32.5^o

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The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

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