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Nostrana [21]
3 years ago
12

A projectile is launched horizontally at off a 20m high position. It goes 60m

Physics
1 answer:
Elenna [48]3 years ago
7 0
The time in flight is
from s= 1/2 g t^2
so t = square root ((2x20)/ 9.8)
t = 2 seconds.

This means the horizontal speed was 30 m/s
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Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

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v is the linear velocity

\omega is the angular velocity

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In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

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r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

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v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

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Explanation:

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Answer:

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