Answer:
201.6 N
Explanation:
m = mass of disk shaped merry-go-round = 125 kg
r = radius of the disk = 1.50 m
w₀ = Initial angular speed = 0 rad/s
w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s
t = time interval = 2 s
α = Angular acceleration
Using the equation
w = w₀ + α t
4.296 = 0 + 2α
α = 2.15 rad/s²
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²
F = constant force applied
Torque equation for the merry-go-round is given as
r F = I α
(1.50) F = (140.625) (2.15)
F = 201.6 N
Vi=12m/s Vf=16m/s t=8s a=? a=Vf-Vi/t=16-12/8=4/8=1/2 a=0.5m/s^2
answer: transverse and longitudinal
Answer:
3600N
Explanation:
Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s
ΣF = ma
we need to find 'a' first, using the definition of 'a' we get equation:
a = (Vf-Vo)/Δt
a = (30m/s)/10s
a = 3 m/s^2
now substitute into top equation
ΣF = ma
Fengine = (1200kg)(3m/s^2)
Fengine = 3600N
