The work to stretch a spring from its rest position is
(1/2) (spring constant) (distance of the stretch)²
E = 1/2 k x² .
You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write
1700 joules = 1/2 k (3m)²
1 joule = 1 newton-meter
1700 N-m = 1/2 k (3m)²
Multiply each side by 2: 3400 N-m = k · 9m²
Divide each side by 9m² k = 3400 N-m / 9m²
= (377 and 7/9) newton per meter
Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
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