The skater's final angular speed is equal to 12 rad/s.
When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.
The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.
In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.
Learn more about angular momentum here brainly.com/question/7538238
#SPJ4
Answer:
force (tension) of 29.4 N (upward) in 100 cm
force (tension) of 58.4 N (upward) in 200 cm
Explanation:
Given:
Length of tube = 5 m (500 cm)
Mass of tube = 9
Suspended vertically from 150 cm and 50 cm.
Computation:
Force = Mass × gravity acceleration.
Force = 9.8 x 9
Force = 88.2 N
So,
Upward forces = Downward forces
D1 = 150 - 50 = 100 cm
D2 = 150 + 50 = 200 cm
And F1 = F2
F1 x D1 = F2 x D2
F1 x 100 = F2 x 200
F = 2F
Total force = Upward forces + Downward forces
3F = 88.2
F = 29.4 and 2F = 58.8 N
force (tension) of 29.4 N (upward) in 100 cm
force (tension) of 58.4 N (upward) in 200 cm
Answer:
A.) 8 m/s
B.) 7.0 m
Explanation:
Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.
(a) What is its velocity when it reaches the top of the plane?
Since the plane is frictionless, the final velocity V will be the same as 8 m/s
The velocity will be 8 m/s as it reaches the top of the plane.
(b) How far horizontally does it land after it leaves the plane?
For frictionless plane,
a = gsinø
Acceleration a = 9.8sin28
Acceleration a = 4.6 m/s^2
Using the third equation of motion
V^2 = U^2 - 2as
Substitute the a and the U into the equation. Where V = 0
0 = 8^2 - 2 × 4.6 × S
9.2S = 64
S = 64/9.2
S = 6.956 m
S = 7.0 m
Answer:
1.71 km
Explanation:
Convert 30 minutes to seconds:
30 min × (60 s / min) = 1800 s
Find the displacement:
0.95 m/s × 1800 s = 1710 m
Convert to kilometers:
1710 m × (1 km / 1000 m) = 1.71 km