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Rainbow [258]
2 years ago
8

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

. Determine the wheel's average rotational acceleration.
Physics
1 answer:
earnstyle [38]2 years ago
3 0

Answer:

0.4rad/s²

Explanation:

Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)

w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m

a = ∆w/t

a = (w2 - w1)/t

a = (2 - 4)/5 = -2/5 =

a = - 0.4rad/s²

The -ve sign indicates a deceleration in the motion

Good luck

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A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha
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Answer:

The efficiency is 0.33, or 33%.

Explanation:

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\eta=1-\frac{200J}{300J}=0.33

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3 years ago
Read 2 more answers
A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu
boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

a_{t} = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

a_{r} = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

a = \sqrt{a_{t}^{2} + a_{r}^{2}}

a = \sqrt{0.18^{2} + 0.38^{2}}

a = 0.42 m/s²

8 0
3 years ago
What must occur for work to be done on an object?
katrin2010 [14]
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Hope this helped! If it did, would you mind marking my answer as Brainliest? Thanks :D
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