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Anni [7]
3 years ago
14

A rock is launched vertically upward by a slingshot. The elastic band of the slingshot accelerates the rock from rest to 40 m/s

in a distance of 10 cm. How long will the rock take to return to its release point (time of flight)? Neglect air resistance.
Physics
1 answer:
eimsori [14]3 years ago
5 0

Answer:

0.016 s

Explanation:

Initial velocity, u = 0

use the third equation of motion to find the acceleration of the rock.

2as=v^2-u^2\\a = \frac{40^2-0}{2\times0.1} =8000 m/s^2

Net displacement in the vertical direction would be zero. y = 0

Use second equation:

s=ut+\frac{1}{2}at^2\\0=0+\frac{1}{2}(8000)t^2\\t= 0.016 s

Thus, the rock will return to its initial release point in 0.016 s.

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If the initial upward speed of the ball in Activity 7 C.2 is 10m/s, and the ball is releasedat a height of 1.5 m above the oor,
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6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.

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5 0
3 years ago
Read 2 more answers
A progressive wave is represented by Y=Asin (wt +pi/6) . If displacement at the origin at time 0 is 3.5cm, if velocity is 500cms
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3 years ago
a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we
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Explanation:

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∑τ = Iα

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∑F = ma

F − 100 N − 200 N = 0

F = 300 N

8 0
4 years ago
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