3.
(a)
ϕ=ω0t+ϵt22=2⋅2+3.5⋅222=11rad\phi=\omega_0t+\frac{\epsilon t^2}{2}=2\cdot2+\frac{3.5\cdot2^2}{2}=11 radϕ=ω0t+2ϵt2=2⋅2+23.5⋅22=11rad
(b)
n=ϕ2π=112π=1.75n=\frac{\phi}{2\pi}=\frac{11}{2\pi}=1.75n=2πϕ=2π11=1.75
(c)
ω=ω0+ϵt=2+3.5⋅2=9rad/s\omega=\omega_0+\epsilon t=2+3.5\cdot2=9rad/sω=ω0+ϵt=2+3.5⋅2=9rad/s
4.
4000rpm≈419rad/s4000rpm\approx419rad/s4000rpm≈419rad/s
ω=ω0−ϵt→0=ω0−ϵt→t=ω0/ϵ=419/20≈21s\omega=\omega_0-\epsilon t\to0=\omega_0-\epsilon t\to t=\omega_0/\epsilon=419/20\approx21sω=ω0−ϵt→0=ω0−ϵt→t=ω0/ϵ=419/20≈21s
ϕ=ω0t−ϵt22=419⋅21−20⋅2122=4389rad\phi=\omega_0t-\frac{\epsilon t^2}{2}=419\cdot21-\frac{20\cdot21^2}{2}=4389 radϕ=ω0t−2ϵt2=419⋅21−220⋅212=4389rad
n=ϕ2π=43892π=698.5n=\frac{\phi}{2\pi}=\frac{4389}{2\pi}=698.5n=2πϕ=2π4389=698.5
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1. A 1.60 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.25 cm when tightened? 2. A load of 50 kg is applied to the lower end of a steel rod 80 cm long and 0.6 cm in diameter. How much will the rod stretch? Y = 190 GPa for steel. 3. An iron rod 4 m long and 0.5 cm2 in cross-section stretches 1 mm when a mass of 225 kg is hung from its lower end. Compute Young's modulus for the iron.
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Dear visitor, please use panel for submitting new questions
1. A 1.60 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.25 cm when tightened? 2. A load of 50 kg is applied to the lower end of a steel rod 80 cm long and 0.6 cm in diameter. How much will the rod stretch? Y = 190 GPa for steel. 3. An iron rod 4 m long and 0.5 cm2 in cross-section stretches 1 mm when a mass of 225 kg is hung from its lower end. Compute Young's modulus for the iron.
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