Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules
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Answer: Volume = 1080m^3
Explanation:
Given that the prism has a 15 m by 18 m rectangular base and a height of 4 m
Volume is the product of length, breath and height. That is
Volume = L × B × H
Where
L = 18 m
B = 15m
H = 4m
Using the formula above gives:
Volume V = 18 × 15 × 4
V = 1080 m^3
Answer:
<u>ω = 1.7 rad/s</u>
Explanation:
Conservation of angular momentum
Assuming the rod is initially hanging vertically at rest.
Initial angular momentum is carried by the bullet only
L = Iω = (mR²)(v/R) = mvR = 0.020(200)(0.7) = 2.8 kg•m²/s
the same angular momentum exists after impact, only the moment of inertia has increased by that of the rod. I = ⅓mR²
2.8 = (⅓(10)(0.7²) + 0.020(0.7²))ω
2.8 = (1.64313333...)ω
ω = 1.70406134...
