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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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Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

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\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

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The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm . The explorer finds that
iren2701 [21]

Answer: g = 10.0 m/s/s

Explanation:

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Equating both definitions for T, we can solve for g, as follows:

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7 0
3 years ago
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