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Aloiza [94]
3 years ago
15

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity c

hanges, each taking place in a time interval 11.3 s . What are the average acceleration in each interval?
1.) At the beginning of the interval the astronaut is moving toward the right along the x-axis at 15.0 m/s, and at the end of the interval she is moving toward the right at 5.30 m/s.
a = ___________-??
2.) At the beginning she is moving toward the left at 5.30 m/s. and at the end she is moving toward the left at 15.0 m/s.
a =_________ ??
3.) At the beginning she is moving toward the right at 15.0 m/s, and at the end she is moving toward the left at 15.0 m/s.
a = _________??
Physics
1 answer:
stiv31 [10]3 years ago
5 0

Answer

given,

time interval = 11.3 s

a) initial velocity, vi = 15 m/s

   final velocity, v_f = -5.30 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-5.30 -15}{11.3}

             a = -1.79 m/s²

   the direction is along left side

b)  initial velocity, vi = -5.30 m/s

   final velocity, v_f = -15 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-15-(-5.30)}{11.3}

             a = -0.858 m/s²

   the direction is along left side

c) initial velocity, vi = 15 m/s

   final velocity, v_f = -15 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-15-(15)}{11.3}

             a = -2.65 m/s²

   the direction is along left side

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4 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

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\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

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