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alisha [4.7K]
3 years ago
8

By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. If the mass of th

e vehicle is 155.1× greater than the astronaut by what factor would the astronaut's acceleration be affected? Think of a method in which recoil of the vehicle is avoided. original text = Discuss how this would affect the measurement of the astronaut’s acceleration.
Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

The astronaut's acceleration is 155.1 times the vehicle's acceleration

Explanation:

These effects due to Newton's third law of action and reaction. Since the forces are equal but in the opposite direction and each acting on a different body. We distance that the Force is F let's calculate the acceleration of the vehicle and the astronaut  

Astronaut

        F = m_{ast}  a₁

Vehicle

        F = m_{veh} a₁

F = 555.1 m_{ast} a₂

Let's match the equation

m_{ast} a₁ = 155.1 m_{ast} a₂

a₁ = 155.1 a₂

a₁ / a₂ = 155.1

The astronaut's acceleration is 155.1 times the vehicle's acceleration

We see that even when the acceleration of the vehicle is small, there is a very high multiplicative factor.

One method to improve this situation is that the vehicles fear some small retro-rocket vehicles to reduce their acceleration. This would have a very favorable impact on the astronaut's mission.

Another method would be for the astronaut himself to have the retro-rocket and control his acceleration.

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Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

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S = 1/2a(t)²

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The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

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  • <em>Length of the pendulum, = L</em>
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The moment of inertia of the rod about the end is given as;

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The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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Answer:

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