Answer:
1606
The main reason for using light years, however, is because the distances we deal with in space are immense. If we stick to miles or kilometers we quickly run into unwieldy numbers just measuring the distance to the nearest star: a dim red dwarf called Proxima Centauri that sits a mere 24,000,000,000,000 miles away!
Explanation:
I think you need to add more.. but I may know where you are leading
Was he 200 m away and made the trip in 200 seconds?
If yes...
2 m/s was his speed and 0 velocity
Answer:
The track's angular velocity is W2 = 4.15 in rpm
Explanation:
Momentum angular can be find
I = m*r^2
P = I*W
So to use the conservation
P1 + P2 = 0
I1*W1 + I2*W2 = 0
Solve to w2 to find the angular velocity
0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2
W2 = 0.435 rad/s
W2 = 4.15 rpm
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
