L = length of the meter stick = 1 m
h = height of center of mass of stick from bottom end on the floor = L/2 = 1/2 = 0.5 m
m = mass of the meter stick
I = moment of inertia of the meter stick about the bottom end
w = angular velocity as it hits the floor
moment of inertia of the meter stick about the bottom end is given as
I = m L²/3
using conservation of energy
rotational kinetic energy of meter stick as it hits the floor = potential energy when it is vertical
(0.5) I w² = m g h
(0.5) (m L²/3) w² = m g h
( L²) w² = 6g h
( 1²) w² = 6 (9.8) (0.5)
w = 5.4 rad/s
Answer:
Because it can desolve many many things.
Explanation:
Answer:
The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
hmax = 5740.48 m
Explanation:
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
Answer:
µ = 
Explanation:
µ = 1/ sinC
µ -----> refractive index of medium
C ----> critical angle
Hope this helps!
The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
Learn more here: brainly.com/question/24430414
