Right, as you mentioned in the comments, you find 
by plugging in the different values of 
.
For 
, we have
Similarly, for 
, you get
Work is (force) times (distance). For Amy, you know both of them, and you can easily multiply them to find the amount of work. For Joe, the distance is zero, which should tell you all you need to know.
Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
Answer:
Could you explain that more better?
Explanation:
<h2>distance = 523 cm</h2>
Explanation:
( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) The definition of frequency is the number of rotations per second .
Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz
( c ) The tangential speed is v = angular velocity x radius of rotation
The angular velocity ω = 2π x n , where n is the number of rotations per second
Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec
The linear velocity = angular velocity x distance from center of record
Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec
Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad
Linear displacement = angular displacement x distance from center of record
= 50π/3 x 10 = 500π/3 = 523 cm
