The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.
i) The angular acceleration of the disc (
), in revolutions per square second, is found by the following kinematic formula:
(1)
Where:
- Initial angular speed, in revolutions per second.
- Final angular speed, in revolutions per second. 
- Time, in seconds.
If we know that 
, 
y 
, then the angular acceleration of the disc is:
The angular acceleration of the disc is 
radians per square second.
ii) The number of rotations that the disk makes before it stops (
), in revolutions, is determined by the following formula:
(2)
If we know that , y , then the number of rotations done by the disc is:
The disc makes 3.125 revolutions before it stops.
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Answer:
0.976 c
Explanation:
= velocity of object 1 relative to earth = 0.80 c
= velocity of object 2 relative to object 1 = 0.80 c
= velocity of object 2 relative to earth
Velocity of object 2 relative to earth is given as
= 0.976 c
Answer:
44 N/m
Explanation:
The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m
The work needed to stretch a spring by <em>e</em> is given by
where <em>k</em> is spring constant.
Using the appropriate values,
Answer:
Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.
