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Romashka-Z-Leto [24]
4 years ago
8

A spring-mounted chair in which the astronaut sits, can be used to find the mass of an astronaut. The chair is then made to osci

llate in simple harmonic motion. The spring used in one such device has a spring constant of 569 N/m, and the mass of the chair is 11 kg. The measured oscillation period is 3.6 s. Find the mass of the astronaut.
Physics
1 answer:
emmainna [20.7K]4 years ago
4 0

Answer:

M = 175 kg

Explanation:

In the resolution of the harmonic oscillator movement of a system and a mass with a spring, the angular velocity is

    w = √ k / m

Where k is the spring constant and m the mass

In this case the mass is the mass of the chair (m) plus the mass of the astronaut (M)

    M all = m + M

The angular velocity and the period are related by

    w = 2π / T

Substituting

   2π / T = √(k/(m + M))

We calculate the astronaut's mass

   4π² / T² = k / (m + M)

   M = k T² / 4π² - m

   M = 569 3.6² /(4π²) - 11

   M = 186.8 - 11

   M = 175 kg

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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