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3 years ago

12

At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration of -0. 40 rad/s2 has an angular velocity of 1.

5 rad/s and an angular position of 2. 3 rad. What is the angular position of the wheel at t = 2. 0 s?.

1 answer:

Novay_Z [31]3 years ago

8 0

The angular position of the wheel at t = 2. 0 s is 4.5 rad.

Angular Position:

When an object is undergoing a rotational motion, then the orientation of the object with respect to a specific reference position is known as angular position.

Given data:

The angular acceleration of the object at initial is, α = - 0.40 rad/s².

The angular velocity of the object is, ω = 1.5 rad/s.

The angular position of the object at initial is, $\theta = 2.3 \;\rm rad$ .

The time interval is, t = 2.0 s.

The expression for the angular position of the wheel at t = 2.0 s is,

$\theta' = \theta+ \omega t + \dfrac{1}{2} \alpha t^{2}$

Solving as,

$\theta' = 2.3+ (1.5 \times 2) + \dfrac{1}{2} \times (-0.40) \times 2^{2}\\\\
\theta' = 4.5 \;\rm rad$

Thus, we can conclude that the angular position of the wheel at t = 2. 0 s is 4.5 rad.

Learn more about the angular position here:

brainly.com/question/6226195

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Answer:

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Explanation:

Given that,

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$\dfrac{dE}{dt}(\pi r^2)=\dfrac{d\phi_{E}}{dt}$

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According to Ampere Maxwell law

$\oint{\vec{B}\cdot \vec{ds}}=\mu_{0}(I+\epsilon_{0}\dfrac{d\phi_{E}}{dt})$

$\oint{\vec{B}\cdot\vec{ds}}=\mu_{0}I+\mu_{0}\epsilon_{0}\dfrac{d\phi_{E}}{dt})$

Electric field is zero inside the circle.

$\oint{\vec{B}\cdot \vec{ds}}=\mu_{0}\epsilon_{0}\dfrac{d\phi_{E}}{dt})$

$B(2\pi\times10.0\times10^{-2})=4\pi\times10^{-7}\times8.85\times10^{-12}\times0.0503$

$B=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times0.0503}{2\pi\times10.0\times10^{-2}}$

$B=8.9\times10^{-19}\ T$

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Answer: 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

Explanation:

1) Since the mass of <span>155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).
</span>

<span>2) You do that using the trignometric ratios sine and cosine.
</span>

<span>Weight is mass × g.
</span>
<span>Weight of the object = 155g × g
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</span>

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3) Conclusion:

Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

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