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.
Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.
Answer:
The skater has mechanical/gravitational potential energy at the two meter mark. The skater gets to two meters high on the other end of the ramp. In terms of the conservation of energy, the skater will never go higher than two meter on the other end of the the ramp because energy can be neither created nor destroyed.
Explanation:
I hoping it is right!!!∪∧∪ ∪ω∪
The resultant speed of the plane is (3) 226 m/s
Why?
We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).
So, calculating we have:
Hence, we have that the resultant speed of the plane is (3) 226 m/s
Have a nice day!
That's the "Atomic Mass" of the atom.
