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Fiesta28 [93]
3 years ago
12

A bicycle tire is spinning counterclockwise at 3.00 rad/s. During a time period Δt = 2.30 s, the tire is stopped and spun in the

opposite (clockwise) direction, also at 3.00 rad/s. Calculate the change in the tire's angular velocity Δω and the tire's averageangular acceleration α. (Indicate the direction with the signs of your answers.)
Physics
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer:

Δω =  -6.00 rad/s

α = -2.61 m/s²

Explanation:

Step 1: Data given

A bicycle tire is spinning counterclockwise at 3.00 rad/s

Δt = 2.30 s

In theopposite (clockwise) direction, also at 3.00 rad/s

Step 2: Calculate the change in the tire's angular velocity Δω

Δω =  ωf -  ωi

ωf = - 3.00 rad/s

 ωi  = 3.00 rad/s

Δω =  ωf -  ωi  = -3.00 - 3.00 = -6.00 rad/s

Step 3: Calculate the tire's average angular acceleration α

α =  Δω /  ΔT

α = -6.00 rad/s /2.30s  

α = -2.61 m/s²

A negative angular acceleration means a decreasing angular velocity

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Answer:

a. cosθ b. E.A

Explanation:

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b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

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Answer

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    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

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