A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The
beam balances when a fulcrum is placed below the beam a distance 1.10 m from the 30.0-kg child. How long is the beam?
1 answer:
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
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Answer:
33.6371 m
Explanation:
t = Time taken
u = Initial velocity = 20.3 m/s
v = Final velocity
s = Displacement
a = Acceleration = -7 m/s²
Distance traveled in the 0.207 seconds
Distance = Speed × Time
⇒Distance = 20.3×0.207 = 4.2021 m
Equation of motion
Distance traveled by the car while braking is 29.435 m
Total distance measured from the point where the driver first notices the red light is 29.435+4.2021 = 33.6371 m