Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
Answer:
25.8 rad/s
Explanation:
C = circumference of the hollow sphere = 0.749 m
r = radius of the sphere
Circumference of the hollow sphere is given as
C = 2π r
0.749 = 2 (3.14) r
r = 0.12 m
m = mass of the basketball = 0.624 kg
Moment of inertia of basketball is given as
I = 5.99 x 10⁻³ kg-m²
w = angular speed
KE = Kinetic energy of the ball = 1.99 J
Kinetic energy of the ball is given as
KE = (0.5) I w²
1.99 = (0.5) (5.99 x 10⁻³) w²
w = 25.8 rad/s
A convex lens<span> makes </span>light<span> rays </span><span>converge </span>
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