I'm going to say false hope that helped
Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
Answer:
(a) 0.017m/s^2
(b) 17/100,000
(c) 0.17m, 0.558ft
Explanation:
(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s
Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2
(b) g = 9.8m/s^2, a = 0.0017m/s^2
a/g = 0.0017/9.8 = 0.00017 = 17/100,000
(c) Distance = speed × time = 0.017m/s × 10s = 0.17m
Distance in foot = 0.17 × 3.2808ft = 0.558ft
Answer:
a.) volume = 70cm^3
b.) mass = 805 grams
Explanation:
see the explanation in attachment.
