Answer
given,
mass of the shuffleboard disk = 0.42 kg
speed of the cue is increased to = 4.2 m/s
acceleration takes over 2 m then acceleration is zero.
the disk additionally slide to 12 m
final speed of disk = 0 m/s
a) increase in thermal energy
b) 
F_f is the frictional force
increase in thermal energy for entire movement of 14 m
c) Work done on the disk by the cue
W = 3.704 + 0.616
W = 4.32 J
True the only way to emit ur proton is if there are ground electrical
Frequency
Amplitude
Wavelength
Speed
Answer:
ω₂ = 93.6 rev / min
Explanation:
ω₀ = 260 rev / min
ω₁ = 0.68*ω₀ = 0.68*(260 rev / min) = 176.8 rev / min
ω₂ = ?
t₁ = 1 min
t₂ = 2 min
We can apply the equation:
ω₁ = ω₀ + α*t₁ ⇒ α = (ω₁ - ω₀) / t₁
⇒ α = (176.8 rev / min - 260 rev / min) / 1 min = - 83.2 rev / min²
then we can use the same formula, knowing the angular acceleration:
ω₂ = ω₀ + α*t₂ ⇒ ω₂ = (260 rev / min) + (- 83.2 rev / min²)*(2 min)
⇒ ω₂ = 93.6 rev / min
