Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
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.
Answer:
Explanation:
Given that, the distance between the electrode is d.
The electron kinetic energy is Ek when the electrode are at distance "d" apart.
So, we want to find the K.E when that are at d/3 distance apart.
K.E = ½mv²
Note: the mass doesn't change, it is only the velocity that change.
Also,
K.E = Work done by the electron
K.E = F × d
K.E = W = ma × d
Let assume that if is constant acceleration
Then, m and a is constant,
Then,
K.E is directly proportional to d
So, as d increase K.E increase and as d decreases K.E decreases.
So,
K.E_1 / d_1 = K.E_2 / d_2
K.E_1 = E_k
d_1 = d
d_2 = d/3
K.E_2 = K.E_1 / d_1 × d_2
K.E_2 = E_k × ⅓d / d
Then,
K.E_2 = ⅓E_k
So, the new kinetic energy is one third of the E_k
Answer:
Check the correctness of the relation c=1 √μ0∈0 where the symbols have their usual meaning. check-circle.