Answer:
Explanation:
Given
mass 
Force 
door knob is located at a distance of r=0.8 m from axis
Angular acceleration of door 
Torque 
where I=moment of inertia
Answer:
it started to move a 1 second
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
Answer:
r=6.05km/hr
z=59.1 degree to the horizontal
Explanation:
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place
can be solved using pythagoras theorem
r^2=o^2+a^2
r^2=5.2^2+3.1^2
r^2=36.65
r=6.1km/hr is te birds velocity relative to the ground
tanz=5.2/3.1
z=tan^-1(5,2/3.1)
z=59.1 degree to the horizontal
