Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J
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Time required : 3 s
<h3>Further explanation
</h3>
Power is the work done/second.
To do 33 J of work with 11 W of power
P = 11 W
W = 33 J
Answer:
(1) V = 0.2 J (2) 0.05J
Explanation:
Solution
Given that:
K = 160 N/m
x = 0.05 m
Now,
(1) we solve for the initial potential energy stored
Thus,
V = 1/2 kx² = 0.5 * 160 * (0.05)²
Therefore V = 0.2 J
(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.
By using the energy conversion, we have the following
ΔV = mgh
=(0.1/9.8) * 9.8 * 1.5 = 0.15J
The internal energy = 0.2 -0.15
=0.05J
Answer:
your in mr langfords class
Explanation:
bruh moment
