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3 years ago

If the thrower takes 0.90 s to complete one revolution, starting from rest, what will be the speed of the discus at release?

1 answer:

8 0

Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled

Let
α = the angular acceleration
ω = the final angular velocity

The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²

The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s

If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s

Answer: 13.963 rad/s

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Strike441 [17]

The solution would be like this for this specific problem:

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Grant distance of each maxima from centre = y ..
As sinθ ≈ y/D y/D = λ/d λ = yd / D 

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3 years ago

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umka2103 [35]

Answer:

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Explanation:

Given:

Magnetic field, B = 4.35 T

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Magnetic flux is determine by the relation:

$\Phi = BA\cos \theta$ ....(1)

Here A represents area of the circular loop.

Area of circular loop, A = πr²

Hence, the equation (1) becomes:

$\Phi=B\pi r^{2} \cos \theta$

Substitute the suitable values in the above equation.

$\Phi=4.35\times\pi (0.28)^{2} \cos 15.1$

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2 years ago

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