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3 years ago

Explain why it is important to create measurable goals.

Physics

2 answers:

ANTONII [103]3 years ago

3 0

Answer:

it is important to create measurable goals bc :

To determine the point at which the goal has been achieved.
It can also help improve motivation allowing an individual to track the progress towards the goal.

hope it helps!

Nat2105 [25]3 years ago

3 0

Answer:

It is important to create measurable goals in order to determine the point at which the goal has been achieved. Also, a measurable goal can help improve motivation by allowing an individual to effectively track progress toward meeting the goal.

Explanation:

sorry i'm late, hope this helps!

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The speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

Further Explanation:

Given:

The angular velocity of the disk is $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The radius of the circular disk is $0.1\,{\text{m}}$ .

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The linear speed of the bug present at the rim of the circular disk is given by.

$v = r \times \omega$

Here, $v$ is the linear speed, $r$ is the radius of the disk and $\omega$ is the angular speed of the disk.

Substitute $0.1\,{\text{m}}$ for $r$ and $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $\omega$ in above expression.

$\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}$

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

${a_c} = \dfrac{{{v^2}}}{r}$

Substitute $1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $v$ and $0.1\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}$

Thus, the speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords: Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

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