Answer:
E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2
E = 1/2 M V^2 (1 + I / (M R^2))
For a cylinder I = M R^2
For a sphere I = 2/3 M R^2
E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2
E(sphere) = 1 + 2/3 = 1.67
E(cylinder) / E(sphere) = 2 / 1.67 = 1.2
The cylinder initially has 1.20 the energy of the sphere
The PE attained is proportional to the initial KE
H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE
The momentum of the ball when it hits the ground is 4.89 kg.m/s.
The given parameters;
- <em>mass of the baseball, m = 0.145 kg</em>
- <em>height of fall of the ball, h = 58 m</em>
The final velocity of the ball when it hits the ground is calculated as follows;

The momentum of the ball when it hits the ground is calculated as follows;
P = mv
P = 0.145 x 33.72
P = 4.89 kg.m/s
Thus, the momentum of the ball when it hits the ground is 4.89 kg.m/s.
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Answer:
the resistors In series have much power
Given: distance 1 d₁ = 40 m; distance 2 d₂ = 3.8 m g = -9.8 m/s²
Initial Velocity Vi = 0 Final Velocity of stone 2 is unknown = ?
Total distance dₓ = d₁ - d₂ = 40 m - 3.8 m = 36.2 m
Formula: a = Vf² - Vi²/2d derive for Final Velocity Vf
acceleration is now due to gravity, therefore a = g
Vf = √2gd Vf = √2(9.8 m/s²)(36.2 m)
Vf = 26.64 m/s
Reason: The second stone will still start from rest.
Answer:
41.8m/s^2
Explanation:
Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s
From the equations of motion, v = u + at
a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2