Work = force × distance
= 35 N × 200 m
= 7000 J
Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay


- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s