Answer:
a) w = 2.57 rad / s
, b) α = 3.3 rad / s²
Explanation:
a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground
Initial. Higher
Em₀ = U = m g h
Final. Touching the ground
= K = ½ I w²
How energy is conserved
Em₀ =
mg h = ½ I w2
The moment of specific object inertia
I = m L²
We replace
m g h = ½ (mL²) w²
w² = 2g h / L²
The height of the object is the length of the bar
h = L
w = √ 2g / L
w = √ (2 9.8 / 2.97)
w = 2.57 rad / s
b) the angular acceleration can be found from Newton's second rotational law
τ = I α
W L = I α
mg L = (m L²) α
α = g / L
α = 9.8 / 2.97
α = 3.3 rad / s²
<span>95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M</span>
Answer:
Explanation:
As the circuit is parallel, then there is no effect of other branches as the potential difference across each arm is same.
Great question the answer is -25x.
