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Crazy boy [7]
4 years ago
10

An object is dropped from 26 feet below the tip of the pinnacle atop a 702 ft tall building. The height h of the object after t

seconds is given by the equation h=-16t^2+676. Find how many seconds pass before the object reaches the ground.
Physics
1 answer:
Dimas [21]4 years ago
4 0

Answer:

6.5 seconds.

Explanation:

Given: h=-16t²+679

When the object reaches the ground, h=0.

∴ 0=-16t²+679

collecting like terms,

⇒ 16t²=679

Dividing both side of the equation by the coefficient of t² i.e 16

⇒ 16t²/16 = 679/16

⇒ t² = 42.25

taking the square root of both side of the equation.

⇒ √t² =√42.25

⇒ t = 6.5 seconds.

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Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

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v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

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Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

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T = 24.9 N

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A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
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Answer:

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The box will stop when it's final velocity becomes zero

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