Answer:
They are all a cycle!
Explanation:
They just are all cycles.
Answer:
Acceleration of the meteorite, 
Explanation:
It is given that,
A Meteorite after striking struck a car, v = 0
Initial speed of the Meteorite, u = 130 m/s
Distance covered by Meteorite, s = 22 cm = 0.22 m
We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :



So, the deceleration of the Meteorite is
. Hence, this is the required solution.
Answer:
Explanation:
We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .
14 m due west = - 14 i
22.0 m upward in the elevator = 22 k
12 m north = 12 j
6.00 m east = 6 i
Total displacement = - 14 i + 22 k + 12 j + 6 i
D = - 8 i + 12 j + 22 k
magnitude = √ ( 8² + 12² + 22² )
= √ ( 64 + 144 + 484 )
= √ 692
= 26.3 m
Net displacement from starting point = 26.3 m .
<span> Second-level consumer </span>
(a) Her distance from the starting location is 21.05 m.
(b) The length of the path she skated is 21.05 m.
<h3>
Distance of the skater from the starting position</h3>
The distance around a complete circular path is calculated as 2πr.
The distance for a half circle is calculated as ¹/₂ x 2πr = πr
Distance from the starting location = π x 6.7 m = 21.05 m
The length of the path she skated is the same as her distance from the starting location = 21.05 m.
Learn more about distance round a circle here: brainly.com/question/3100527
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