Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
Answer:
I₁ > I₃ > I₂
Explanation:
Taking the pic shown, we have
m₁ = 10m₀
m₂ = 2m₀
m₃ = m₀
r₁ = r₀
r₂ = 2r₀
r₃ = 3r₀
We apply the formula
I = mr²
then
I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²
I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²
I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²
finally we have
I₁ > I₃ > I₂
If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas
