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coldgirl [10]
3 years ago
7

The angular quantities: angular displacement, angular velocity, angular acceleration, are defined in a very similar way to the l

inear quantities. The angular displacement represents the angle through which an object has rotated (the difference between two angular positions). The angular velocity equals the angular displacement divided by the elapsed time. The angular acceleration equals the change in the angular velocity divided by the elapsed time. Suppose an object travels through an angular displacement, Δθ, of 50.9 rad over a time of 4.10 s. What is the average angular velocity, ω?
Physics
1 answer:
Lena [83]3 years ago
4 0

Answer: [tex]12.415 rad.s^{-1}[/tex]

Explanation: Angular velocity is the rate of change in angular displacement.

We know that:

Angular velocity,      \omega= \frac{\Delta \theta}{t}....................(1)

where:

  • t= time
  • \Delta \theta = angular displacement in radians

<u>Given that:</u>

  • t = 4.10 s
  • Δθ = 50.9 radian

Putting the respective values in eq. (1)

\omega = \frac{50.9}{4.10}

\omega = 12.415 rad.s^{-1}

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What is the momentum of a 12 kg condor flying at 6 m/s?
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Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin
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A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

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