Which derived quantity might be measured in mm^3
1 answer:
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Answer:
a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis
0.7 kg is 109.4 ms on the x axis
b) Y = 109.3 m
Explanation:
This is a moment and projectile launch exercise.
a) Let's start by finding the initial velocity of the projectile
sin 40 = voy / v₀
= v₀ sin 40
= 50.0 sin40
= 32.14 m / s
cos 40 = v₀ₓ / V₀
v₀ₓ = v₀ cos 40
v₀ₓ = 50.0 cos 40
v₀ₓ = 38.3 m / s
Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis
Let's write the amounts
Initial mass of the projectile M = 2.0 kg
Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and = -10.0 m / s
Fragment mass 2 m₂ = 0.7 kg moves in the x direction
Fragment mass 3 m₃ = 0.3 kg moves up (y axis)
Moment before the break
X axis
p₀ₓ = m v₀ₓ
Y Axis y
= 0
After the break
X axis
= m₂ v₂
Axis y
= m₁ v₁ + m₃ v₃
Let's write the conservation of the moment and calculate
Y Axis
0 = m₁ v₁ + m₃ v₃
Let's clear the speed of fragment 3
v₃ = - m₁ v₁ / m₃
v₃ = - (-10) 1 / 0.3
v₃ = 33.3 m / s
X axis
M v₀ₓ = m₂ v₂
v₂ = v₀ₓ M / m₂
v₂ = 38.3 2 / 0.7
v₂ = 109.4 m / s
The fragment speeds of 0.3 kg is 33.3 m / s on the y axis
0.7 kg is 109.4 ms on the x axis
b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics
² =
² - 2 gy
0 = ²-2gy
y = ² / 2g
y = 32.14² / 2 9.8
y = 52.7 m
This is the height where the break occurs, which is the initial height for body movement of 0.3 kg
² =
² - 2 g y₂
0 = ² - 2 g y₂
y₂ = ² / 2g
y₂ = 33.3²/2 9.8
y₂ = 56.58 m
Total body height is
Y = y + y₂
Y = 52.7 + 56.58
Y = 109.3 m
Answer:
New moment of inertia will be
Explanation:
It is given initially angular velocity
Moment of inertia
Angular momentum is equal to
Now angular velocity is decreases to
As we know that angular momentum is conserved
So
So new moment of inertia will be