Answer:
Explanation:
given,
mass of wheel(M) = 3 Kg
radius(r) = 35 cm
revolution (ω_i)= 800 rev/s
mass (m)= 1.1 Kg
I_{wheel} = Mr²
when mass attached at the edge
I' = Mr² + mr²
using conservation of angular momentum
Explanation:
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Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
Answer:
3.06 m/s²
Explanation:
Acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (59.134 m/s − 31.9 m/s) / 8.9 s
a = 3.06 m/s²
