First, we determine the volume of the trunk by finding
first the radius from the circumference through the equation,
<span> C
= 2πr</span>
<span> r
= C/2π</span>
Substituting the known values,
<span> r
= 4.5/2π = 0.716 m</span>
Then, we calculate for the volume through the equation,
<span> V
= πr2h</span>
<span> V
= π(0.716 m)2(8m) = 12.9 m3</span>
Multiplying the calculated value to the density will give
the mass as,
<span> Mass
= (12.9 m3)(752 kg/m3) = <span>9699.36 kg</span></span>
The distance from the Earth to the Sun is 92.96 million mi.
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.
<span>Kinetic Energy means energy that a body possesses by virtue of being in motion. I hope this helps!!!</span>
Friction and the applied force are not balanced. The 7 N push is countered by the frictional force of 3 N to push the box with a net force of 4 N.
