<h2>The time taken is 1 second </h2>
Explanation:
Suppose the first diamond falls for t seconds
The downward displacement S₁ = u t + 
g t²
here u = 0 , because it was at rest initially
Thus S₁ = g t² I
Similarly second diamond moves for time ( t - 0.9 ) sec
The distance covered by it is
S₂ = g (t - 0.9 )²
The separation between the two is 13 m
Thus S₁ - S₂ = 13 m
g t² - g t² + 0.9 g t - 0.4 g = 13
or 9 t = 13 - 4 = 9
Hence t = 1 sec
True
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Answer:
a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²
Explanation:
You may or may not need the negative sign, depending on how the question designer was thinking about the problem.
Since we're dealing with radial acceleration around a circle, I used the radial acceleration equation a=v²/r. At the top of the hill, the force upward exerted by the hill is less than the weight of the sled. if v is large enough the term (g-v²/r) will become 0 and the sled will fly off the ground as it reaches the peak. Let me know if I can clarify any of my work.
