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
1N=1kg•m/s^2 so the answer is 3N
For any object thrown upwards where only the force of gravity is acting upon it, uses the following formula for the maximum height attained.
H= v²/2g, where g = 9.81 m/s²
There are two information of velocities are given. However, we use the 20 m/s information because this is the launch velocity. Hence, the solution is as follows:
H = (20 m/s)²/2(9.81 m/s²)
<em>H = 20.4 m</em>
Answer:
c) 100,000 m/s
Explanation:
You need to take the same wave length from the top graph and bottom one, so let's take half a wave length then in the top one that is 0.005, but in the bottom one it's 2000/4 = 500 because they are smaller and there are 4 half waves before you get to 2000, whereas in the top one there is 1 half wave before you get to 0.005 on the graph.
Now use speed = distance / time
speed = 500 / 0.005 = 100 000 m/s
