The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
<h2>
Answer:</h2>
Motor
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Explanation:</h2>
A motor is a machine that converts electrical energy into mechanical energy. In motors, electric energy is converted into mechanic energy when a magnetic torque acts on a conductor that carries a current. There are different types of motors like DC and AC motors. The moving part of a motor is called the rotor while the stationary part is called stator
Answer:
Explanation:
Given:
- thickness of the base of the kettle,
- radius of the base of the kettle,
- temperature of the top surface of the kettle base,
- rate of heat transfer through the kettle to boil water,
- We have the latent heat vaporization of water,
- and thermal conductivity of aluminium,
<u>So, the heat rate:</u>
<u>From the Fourier's law of conduction we have:</u>
where:
area of the surface through which conduction occurs
temperature of the bottom surface
is the temperature of the bottom of the base surface of the kettle.
Equation 1 :
m1 : m1a =T
Equation 2:m2 : m2a= F - T
Adding 1 and 2a=0.073
Placing in equation 1
we get T = 218.18N
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