Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
The moment of inertia of the flywheel is 2.63 kg-
It is given that,
The maximum energy stored on the flywheel is given as
E=3.7MJ= 3.7×
J
Angular velocity of the flywheel is 16000
= 1675.51
So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :
E = 
By rearranging the equation:
I = 
I = 2.63 kg-
Thus the moment of inertia of the flywheel is 2.63 kg-.
Learn more about moment of inertia here;
brainly.com/question/13449336
#SPJ4
Answer:1). Distance of far point x=0.9m
Therefore, since the image is virtual
-f=-x = -0.9m
Power of the concave lenses = 1/f = 1/-0.9
= -1.11D
2 ) near point is 21cm = 0.21m
Power = 4-1/near point
= 4/0.21
= 14.2D.
If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
so s=(0×8)+ (0.5×-10×64)
s=0+(32×-10)
s=32×-10
s=-320metres
