WILL GIVE 5 STARS!!!! HELP ASAP!!!!
1 answer:
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Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):
Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:
Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:
Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration directed downward, initial vertical position
, and initial vertical velocity
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
substituting, we find
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
In order to do that, we use again the same SUVAT equation substituting with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
Which means that the heigth of the packet was