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rjkz [21]
3 years ago
15

A disk with mass m = 9.5 kg and radius r = 0.3 m begins at rest and accelerates uniformly for t = 18.1 s, to a final angular spe

ed of ω = 28 rad/s. 1) what is the angular acceleration of the disk? rad/s2 2) what is the angular displacement over the 18.1 s? rad 3) what is the moment of inertia of the disk? kg-m2 4) what is the change in rotational energy of the disk? j 5) what is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 6) what is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 7) what is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s 8) what is the total distance a point on the rim of the disk travels during the 18.1 seconds?
Physics
1 answer:
Eddi Din [679]3 years ago
8 0
Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²

2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads

3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²

4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J

5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²

6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²

7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s

8. s = Θ r = 266 rads * 0.33m = 88 m
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A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

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