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Nikitich [7]
3 years ago
9

You want to see how your new belt buckle matches your new hat. Your eyes are 114 mm below the top of the hat and 800 mm above th

e buckle. You walk up to your mirror and notice that the mirror is exactly large enough and exactly in the right position for you see both the top of the hat and the buckle.
1. What is the position of the top edge of the mirror relative to your eyes?2. How tall is the mirror?
Physics
1 answer:
damaskus [11]3 years ago
7 0

Answer:

x=0.057m

h=0.457m

Explanation:

From the question we are told that

Distance of eyes from the top of hat d_1=114mm

Distance of eyes from buckle  d_2=800mm

Generally position of mirror relative eyes x is mathematically given as

x=\frac{114}{2}

x=57mm

x=0.057m

Generally the height of the mirror h is mathematically given as

h=\frac{(Buckle\ below\ the\ hat)+(Buckle\ below\ the\ hat)}{2}

h=\frac{(114)+(800)}{2}

h=457mm

h=0.457m

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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}}}}}}.

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

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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}  

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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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