Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
<u></u>
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
Explanation:
Take south to be negative.
a. Momentum is mass times velocity.
p = mv
p = (540 kg) (-6 m/s)
p = -3240 kg m/s
p = 3240 kg m/s south
b. Impulse = change in momentum
J = Δp
Since the mass is constant:
J = mΔv
J = (540 kg) (-4 m/s − (-6 m/s))
J = 1080 kg m/s
J = 1080 kg m/s north
"Constant velocity" is another way of saying "zero acceleration".
Gamma rays<span> are </span>produced<span> in the disintegration of radioactive atomic nuclei and in the decay of certain subatomic particles. When an unstable atomic nucleus decays into a more stable nucleus, the “daughter” nucleus is sometimes </span>produced<span> in an excited state.</span>
I think its 13...........
