Answer:
C.) 1.5 kg
Explanation:
Start with the equation:

Plug in what you know, and solve:

Find matching soluation:
C.) 1.5 kg
A uniform thin solid door has height 2.20 m, width .870 m, and mass 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unnecessary? So far, I don't understand how to calculate moments of inertia for things like this at all. I can do a system of particles, but when it comes to any ridgid objects, such as this door or rods or cylinders, I don't get it. So basically I have no idea where to even start with this.
so A
Condensation. Remember, Vaporization happens when energy is taken in (enfothermic) the opposite will be the process that releases energy ( exothermic) which will be condensation. Put ice in a glass of water. the ice melts, taking in energy from the water in the glass, which in turn takes heat energy away from the vapor in the surrounding air, thus causing the water vapor in the air to condense.
Orbital periods of stars in the Galaxy
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