Answer:
ω₂=1.20
Explanation:
Given that
mass of the turn table ,M= 15 kg
mass of the ice ,m= 9 kg
radius ,r= 25 cm
Initial angular speed ,ω₁ = 0.75 rad/s
Initial mass moment of inertia
Final mass moment of inertia
Lets take final speed of the turn table after ice evaporated =ω₂ rad/s
Now by conservation angular momentum
I₁ ω₁ =ω₂ I₂
ω₂=1.20
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Answer:
The angle formed of the rope with the surface = 40°
Force applied = 125Newtons
The displacement covered by the box =25metres
W= FDcos theta
[125×40×cos(40°) ] Joules
= [ (3125×0.76604444311)]Joules
= 2393.88888472 joules(ans)
Hope it helps
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Answer:
it relates to the light propensity to travel over one straight line without having any interference in its trajectory
Explanation: