Answer:
20.62361 rad/s
489.81804 J
Explanation:
= Initial moment of inertia = 9.3 kgm²
= Final moment of inertia = 5.1 kgm²
= Initial angular speed = 1.8 rev/s
= Final angular speed
As the angular momentum of the system is conserved
The resulting angular speed of the platform is 20.62361 rad/s
Change in kinetic energy is given by
The change in kinetic energy of the system is 489.81804 J
As the work was done to move the weight in there was an increase in kinetic energy
Try 40. it seems correct. i’m sorry if i’m wrong.
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
Answer:
x = 76.5 m
Explanation:
Let's use Newton's second law at the point of contact between the wheel and the floor.
fr = m a
fr = miy N
N-W = 0
N = W
μ mg = m a
a = miu g
a = 0.600 9.8
a = 5.88 m / s²
Having the acceleration we can use the kinematic relationships to find the distance
² = v₀² + 2 a x
= 0
x = -v₀² / 2 a
Acceleration opposes the movement by which negative
x = - 30²/2 (-5.88)
x = 76.5 m
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
