2nd and third laws of energy
Answer:
The answer is below
Explanation:
The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm
radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m
a) Initial angular velocity () = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s
θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
angular acceleration (α) is:
b)
c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
a) When it stops, the final angular velocity is 0. Hence:
θ = 323 rad
Answer:
option D) -3m
Explanation:
if 6m is diplaced by -3m then it would be -3+6=3m
feel free to ask if you are confused
part 1
mass = ρ x V
mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg
PE (potential energy)= mgh
PE = 6608.2 kg x 9.81 x 403
PE = 2.61 x 10⁷ J
part 2
megaton of TNT (Mt) =4.2 x 10¹⁵ J
convert PE to Mt:
2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt
Work Done = Force x Distance Moved
Work Done = 25 x 15 = 375 Joules