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navik [9.2K]
4 years ago
13

NASA operates a 2.2-second drop tower at the Glenn Research Centre in Cleveland, Ohio. At this facility, experimental packages a

re dropped from the top of the tower, on the 8th floor of the building. During their 2.2 seconds of free fall, experiments experience a microgravity environment similar to that of a spacecraft in orbit,
(a) What is the drop distance of a 2.2-s tower?
(b) How fast are the experiments travelling when they hit the air bags at the bottom of the tower?
(c) If the experimental package comes to rest over a distance of 0.75 m upon hitting the air bags, what is the average stopping acceleration?
Physics
1 answer:
alukav5142 [94]4 years ago
3 0

Answer:

23.7402 m

21.582 m/s

310.521816 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m

The drop distance is 23.7402 m

v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s

When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s

The final speed of the fall will be the initial velocity of stopping

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2

The average stopping acceleration is 310.521816 m/s²

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A) The energy stored in a capacitor is given by (1/2) (CV²)

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C) U₂ = CV²/2

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U₂ = ϵAV²/2d

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