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3 years ago

8

A bug is on the rim of a disk of diameter 9 in that moves from rest to an angular speed of 76 rev/min in 6.5 s. what is the tang

ential acceleration?

1 answer:

ehidna [41]3 years ago

5 0

Let's convert the final angular speed of the disk into rad/s:
$\omega_f = 76 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/min}=7.96 rad/s$
while the initial angular speed is zero.

So, the angular acceleration of the disk is
$\alpha = \frac{\omega_f - \omega_i }{t} = \frac{7.96 rad/s}{6.5 s}=1.22 rad/s^2$

The radius of the disk is half the diameter: $r= d/2 = 9 m/2 =4.5 m$

And so, the tangential acceleration is the angular acceleration times the radius:
$a_t = \alpha r =(1.22 rad/s^2)(4.5 m)=5.49 m/s^2$

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Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The eyepiece focal length is $f_e = 0.027 \ m$

Explanation:

From the question we are told that

The focal length is $f_o = 5.5 \ mm = -0.0055 \ m$

This negative sign shows the the microscope is diverging light

The angular magnification is $m = 300$

The distance between the objective and the eyepieces lenses is $Z = 19 \ cm = 0.19 \ m$

Generally the magnification is mathematically represented as

$m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]$

Where $f_e$ is the eyepiece focal length of the microscope

Now making $f_e$ the subject of the formula

$f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }$

substituting values

$f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }$

$f_e = 0.027 \ m$

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Charge on disk

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Note: Refer the image attached

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3 years ago

A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif

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Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

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$sin^{-1}(\frac{gR}{v^{2}})=2\theta$

$\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}$

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θ=0.05614°

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