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sveticcg [70]
4 years ago
12

An object's potential energy is set it cannot change is this true

Physics
2 answers:
zepelin [54]4 years ago
8 0
The energy achieved I think
beks73 [17]4 years ago
4 0
The energy is archived, I think
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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has
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Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration (a) consist in two components, <em>radial</em> (a_{r}) and <em>tangential</em> (a_{t}), in meters per square second:

a_{r} = \omega^{2}\cdot r (1)

a_{t} = \alpha \cdot r (2)

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a = \sqrt{a_{r}^{2}+a_{t}^{2}} (3)

Where:

r - Radius of the wheel, in meters.

\omega - Angular speed, in radians per second.

\alpha - Angular acceleration, in radians per square second.

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\omega = \omega_{o}+ \alpha\cdot t (4)

Where:

\omega_{o} - Initial angular speed, in radians per second.

t - Time, in seconds.

If we know that r = 0.1\,m, \alpha = 2\,\frac{rad}{s^{2}}, \omega_{o} = 0\,\frac{rad}{s} and t = 0.60\,s, then the total linear acceleration is:

\omega = \omega_{o}+ \alpha\cdot t

\omega = 1.2\,\frac{rad}{s}

a_{r} = \omega^{2}\cdot r

a_{r} = 0.144\,\frac{m}{s^{2}}

a_{t} = \alpha \cdot r

a_{t} = 0.2\,\frac{m}{s^{2}}

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

a \approx 0.246\,\frac{m}{s^{2}}

The total linear acceleration is approximately 0.246 meters per square second.

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