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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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Answer:

Reflects

Explanation:

A metal spoon looks very shiny because its surface is very smooth, which is why it acts like a mirror and reflects light very well.

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Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55
masha68 [24]

Answer:

1. X_{cm}=-8.57m

2. Y_{cm}=-14.29m

3. V_{cm} = 16.66m/s

4. α = 59.05°

5. V_{cm} = 16.66m/s

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}

where

X_c=-20m;  m_c=1500kg;  

X_t=0m;   m_t=2000kg;

Replacing these values:

X_{cm}=-8.57m

Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}

where

Y_c=0m;  m_c=1500kg;  

Y_t=-25m;   m_t=2000kg;

Replacing these values:

Y_{cm}=-14.29m

The velocity of their center of mass is:

V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}

where

V_{c-x}=20m/s;  m_c=1500kg;  

V_{t-x}=0m/s;   m_t=2000kg;

Replacing these values:

V_{cm-x}=8.57m/s

V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}

where

V_{c-y}=0m;  m_c=1500kg;  

V_{t-y}=-25m;   m_t=2000kg;

Replacing these values:

V_{cm-y}=-14.29m

So, the magnitude of the velocity is:

V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}

V_{cm}=16.66m/s

The angle of the velocity is:

\alpha =atan(V_{cm-y}/V_{cm-x})

\alpha=59.05\°

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

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Answer:

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swat32

Answer:

4.24m/s

Explanation:

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But

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giving

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