Answer:
(a) 110 rev/ min
(b) 5/6
Explanation:
As per the conservation of linear momentum,
L ( initial ) = L ( final )
I' ω' = ( I' + I'' ) ωf
I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.
(a)
So, ωf = I' ω' / ( I' + I'' )
As I'' = 5I'
ωf = I' ω' / ( I' + 5I' )
ωf = ω'/ 6
now we know ω' = 660 rev / min
therefore ωf = 660/6
= 110 rev/ min
(b)
Initial kinetic energy will be K'
K' = I'ω'² / 2
and final K.E. will be K'' = ( I' + I'' )ωf² / 2
K'' = ( I' + 5I' ) (ω'/ 6)²/ 2
K'' = 6I' ω'²/72
K'' = I' ω'²/ 12
therefore the fraction lost is
ΔK/K' = ( K' - K'' ) / K'
= {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)
= 5/6
When you work out it helps clear your thoughts and think of nothing. It makes you feel good.
Molly & Caden have a stressed
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A =
Let the height of the bin be 'h'
Therefore the total area,
The cost is:
C = 2sh
Volume of the box, V = (1)
Total cost, (2)
From eqn (1):
Using the above value in eqn (1):
Differentiating the above eqn w.r.t 'a':
For the required solution equating the above eqn to zero:
a = 4
Also
The path in order to minimize the cost must be a rectangle.