Answer:
When a neutral atom looses an electron to another neutral atom, two charged atoms are created.
Explanation:
On the off chance that one of the two unbiased particle looses an electron, it turns out to be emphatically (charge: +1), in light of the fact that the electron conveys a charge of - 1. Thus, the other atom which acknowledges the electron turns out to be adversely charged (charge: - 1). And in the end, we will have two charged atoms.
Answer:
After the colision, the stationary electron's momentum is given as:
P = 2.7328 x 10^(-25) kg m/s
The direction of momentum is the same as the direction of velocity of the electron.
Explanation:
In an Isolated system, when an object moving at some velocity v collides head on with a stationary object of equal mass. There velocities are exchanged.
This means that the first electron will become stationary and the electron which was stationary initially will start moving at a velocity of 3*10^(5)m/s in the same direction as the first electron.
Post collision momentum of the stationary electron:
V = 3 x 10^5 m/s
m = 9.1093 x 10^(-31) kg
Momentum = P = mV = 9.1093 x 10^(-31) x 3 x 10^5
P = 2.7328 x 10^(-25) kg m/s
The direction of momentum is the same as the velocity of the electron.
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
