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Sunny_sXe [5.5K]
2 years ago
12

Arrange the following decisions accordingly. Be guided by the chronological circumstances happened from the start of the game. U

se Roman Numerals in answering. 11. Hinto! Hiwalay! Hatol. Unang puntos, Bughaw! 12. Hinto! Pula, pangalawang laglag. Panalo, Bughaw! 13. Hinto! Hiwalay! Bughaw, 1 puntos. Unang paglabag! 14. Hinto! Bughaw, pangalawang paglabag! 15. Hinto! Pula, babala! 16. Hinto! Pula, unang laglag. 17. Hinto! Hiwalay! Hatol. Pula, I puntos. Bughaw, pangatlong paglabag! Panalo, Pula! 18. Hinto! Hiwalay! Bughaw, mag-ayos.​
Physics
1 answer:
Snezhnost [94]2 years ago
3 0

Answer: Use Roman Numerals in answering. 11. - 27310189. ... Hiwalay! Hatol. Unang puntos, Bughaw! 12. Hinto! Pula, pangalawang laglag. Panalo ...

Explanation:. Hinto! Hiwalay! Bughaw, 1 puntos. Unang paglabag! 14. Hinto! Bughaw, pangalawang paglabag

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A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of
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b) 13.12 seconds

c) 2.99 m/s²

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a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2

Emergency acceleration is 2.99 m/s² (magnitude)

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Explanation:

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Explanation:

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