Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
Kinetic energy = (1/2) (mass) (speed)²
= (1/2) (1.4 kg) (22.5 m/s)²
= (0.7 kg) (506.25 m²/s² )
= 354.375 kg-m²/s² = 354.375 joules .
This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.
Here
- Acceleration and initial velocities are constant.
According to first equation of kinematics.
- Time was t at velocity v
- Time will be 4t at velocity 4v
