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
To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:
The angular displacement is given as the form:
In the equlibrium we have to 
and in the given position we have to
Derived the expression we will have the equivalent to angular velocity
Replacing,
Finally
Therefore the maximum angular displacement is 9.848°
Answer:
d = 375 m
Explanation:
The speed of sound is constant in any medium, therefore we can use the uniform motion relationships
v = x / t
x = v t
In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance
x = 2d
2d = v t
d = v t/2
let's calculate
d = 1500 0.5 / 2
d = 375 m
Answer:
The ball has an acceleration of -380 m/s², this means the ball slows down
An acceleration of -380 m/s² is the equivalent of 38.736 g's
Explanation:
Step 1: Data given
Velocity of the baseball at time t=0 = 38 m/s
At time t, the ball stops. This means v = 0
time before stops = 0.1s
Step 2: Calculate the acceleration
v= v0+at
with v= the velocity of the ball at time t = 0. v= 0
with v0 = the velocity of the ball at time t=0. v0 = 38 m/s
with a= the acceleration in m/s²
with t = time in seconds
0 = 38 + a*0.1
a = -380 m/s²
The ball has an acceleration of -380 m/s², this means the ball slows down
An acceleration of -380 m/s² is the equivalent of 38.736 g's
