Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
(I was only able to do A and B)
The answer is c
Explanation:
Answer:
Angular momentum = 0.7 kg.m²/s
Angular velocity = 583.3 rad/s
Explanation:
1. The torque τ is related to the angular momentum L by the relation
τ = ΔL/Δt
ΔL = τΔt
τ = 10 N. m
Δt = 70 ms = 70 × 10⁻³s
ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s
2. The rotational inertia I relates the angular momentum L to the angular velocity w
L = Iw
w = L/I
L = 0.7 kg.m²/s
I = 1.2 × 10⁻³ kg.m²
w = (0.7 kg.m²/s)/(1.2 × 10⁻³ kg.m²) = 583.3 rad/s
C is the answer hope that helps you
